ML Aggarwal Class 6 Maths Solutions Chapter 09 Algebra

 

Exercise 9.1

 

Question 1. Find the rule which gives the number of matchsticks required to make the following matchstick patterns. Use a variable (literal) to write the rule.
(i) A pattern of letter T as
(ii) A pattern of letter V as
(iii) A pattern of letter Z as
(iv) A pattern of letter U as
Answer: Let n denote the number of letters formed.
(i) The letter T needs 2 matchsticks - one placed across the top and one going down the middle. So, to make n T's, the number of matchsticks = 2n.
(ii) The letter V needs 2 matchsticks - two slanted sticks that meet at a point. So, to make n V's, the number of matchsticks = 2n.
(iii) The letter Z needs 3 matchsticks - one at the top going sideways, one going diagonally, and one at the bottom going sideways. So, to make n Z's, the number of matchsticks = 3n.
(iv) The letter U needs 3 matchsticks - one going down on the left side, one going across at the bottom, and one going up on the right side. So, to make n U's, the number of matchsticks = 3n.
In simple words: Count how many matchsticks make one letter, then multiply by n. Each new letter of the same type always needs the same number of sticks.

Exam Tip: Identify the pattern by counting sticks in one unit, then write a formula using n. Always test with n = 1 to check your rule is correct.

 

Question 2. If there are 24 mangoes in a box, how will you write the number of mangoes in terms of the number of boxes? (Use b for the number of boxes.)
Answer: Each box holds 24 mangoes. When you have b boxes, you multiply the mangoes per box by the total number of boxes. So, the number of mangoes in b boxes = 24 × b = 24b.
In simple words: If one box has 24 mangoes, then b boxes will have 24b mangoes.

Exam Tip: When a quantity repeats, multiply the single unit value by the variable representing the count of units.

 

Question 3. Anuradha is drawing a dot Rangoli (a beautiful pattern of lines joining dots). She has 8 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 12 rows?
Answer: In a single row, there are 8 dots. When Anuradha arranges r rows, each with the same number of dots, the total will be 8 × r = 8r dots. If she makes 12 rows, then the total number of dots = 8 × 12 = 96.
In simple words: One row has 8 dots. For r rows, multiply 8 by r. For 12 rows, you get 96 dots.

Exam Tip: Always substitute the given value at the end to find the final numerical answer after writing the algebraic expression.

 

Question 4. Anu and Meenu are sisters. Anu is 5 years younger than Meenu. Can you write Anu's age in terms of Meenu's age? Take Meenu's age as x years.
Answer: Meenu's age is x years. Since Anu is 5 years younger, we subtract 5 from Meenu's age to get Anu's age. Therefore, Anu's age = (x - 5) years.
In simple words: When someone is younger by a fixed number, subtract that number from the older person's age.

Exam Tip: "Younger" means subtraction; "older" means addition. Always match the operation to the relationship described.

 

Question 5. Oranges are to be transferred from larger boxes to smaller boxes. When a larger box is emptied, the oranges from it fill 3 smaller boxes and still 7 oranges are left. If the number of oranges in a smaller box are taken to be x, then what is the number of oranges in the larger box?
Answer: Each small box contains x oranges. Three small boxes will contain 3x oranges. After filling 3 small boxes, 7 oranges remain. Adding the remainder to the amount that fills the smaller boxes gives the total in the larger box: 3x + 7.
In simple words: The big box holds enough to fill 3 small boxes and leave 7 extra. So the big box has 3x + 7 oranges.

Exam Tip: Build expressions step-by-step: identify what happens first (fill boxes), then add the leftover amount.

 

Question 6. Harsha's score in Mathematics is 15 more than three-fourth of her score in Science. If she scores x marks in Science, find her score in Mathematics?
Answer: Harsha's Science score = x marks. Three-fourths of her Science score = \( \frac{3}{4}x \). Since her Math score is 15 more than this amount, we add 15 to get: Mathematics score = \( \frac{3}{4}x + 15 \). Hence, Harsha's score in Mathematics = \( \left(\frac{3}{4}x + 15\right) \) marks.
In simple words: Take three-fourths of the Science score, then add 15 to find the Math score.

Exam Tip: For "more than," add the extra amount. For "times," multiply. Always write fractions in their simplest form.

 

Question 7. Look at the following matchstick pattern of equilateral triangles. The triangles are not separate. Two neighbouring triangles have a common matchstick. Observe the pattern and find the rule that gives the number of matchsticks.
Answer: Let x be the number of triangles. For 1 triangle, we need 3 matchsticks. Each time we add another triangle next to the previous one, that new triangle shares 1 matchstick with its neighbour, so we only add 2 more matchsticks. For x triangles, the total = 3 + 2(x - 1) = 3 + 2x - 2 = 2x + 1. Hence, the rule is (2x + 1) matchsticks for x triangles.
In simple words: The first triangle uses 3 sticks. Every triangle after that adds only 2 more sticks (because one is shared). So x triangles need 2x + 1 sticks.

Exam Tip: When units connect and share resources, count the first unit completely, then add the reduced amount for each additional unit.

 

Question 8. Observe at the following matchstick pattern of letter A. The A's are not separate. Two neighbouring A's have two common matchsticks. Find the rule that gives the number of matchsticks.
Answer: Let x be the number of A's formed. The first A uses 6 matchsticks. The remaining (x - 1) A's each contribute additional sticks; since each new A shares 2 matchsticks with the previous one, each adds 4 new matchsticks. So, for x A's, the total = 6 + 4(x - 1) = 6 + 4x - 4 = 4x + 2. Hence, the rule is (4x + 2) matchsticks for x A's.
In simple words: The first A needs 6 sticks. Each extra A adds 4 more sticks (two are shared with the one before). So x A's need 4x + 2 sticks total.

Exam Tip: Always identify what the first unit uses, then find how much each subsequent unit adds by accounting for shared parts.

 

Exercise 9.2

 

Question 1. If the side of an equilateral triangle is l, then express the perimeter of the triangle in terms of l.
Answer: An equilateral triangle has all 3 sides of equal length l. Perimeter means the total distance around the triangle, so we add all three sides: Perimeter = l + l + l = 3l.
In simple words: All three sides are the same length. Add them up: l + l + l = 3l.

Exam Tip: For perimeter, always add all the side lengths. For equilateral shapes, multiply one side by the number of sides.

 

Question 2. The side of a regular hexagon is l. Express its perimeter in terms of l.
Answer: A regular hexagon has 6 equal sides, each of length l. The perimeter is the sum of all the sides: Perimeter = l + l + l + l + l + l = 6l.
In simple words: A hexagon has 6 sides. Add them all: 6 times l equals 6l.

Exam Tip: For any regular polygon, multiply the side length by the number of sides to find the perimeter.

 

Question 3. The length of an edge of a cube is l. Find the formula for the sum of lengths of all the edges of the cube.
Answer: A cube contains 12 edges, and all of them have the same length l. To find the sum, multiply the number of edges by the length of each: Sum of lengths of all edges = 12 × l = 12l.
In simple words: A cube has 12 edges. Each edge is l long. Multiply: 12 times l is 12l.

Exam Tip: Remember that a cube has 12 edges, a rectangular box (cuboid) also has 12 edges. Always count or recall the number of edges before writing the formula.

 

Question 4. If the radius of a circle is r units, then express the length of a diameter of the circle in terms of r.
Answer: The diameter of a circle is always twice as long as the radius. So, if the radius = r units, then Diameter = 2 × r = 2r units.
In simple words: The diameter is 2 times the radius. So diameter = 2r.

Exam Tip: The relationship between diameter and radius is fixed: Diameter = 2 × Radius. This is a key fact to memorize and apply instantly.

 

Exercise 9.3

 

Question 1. Write all the terms of each of the following algebraic expressions:
(i) 3 - 7x
(ii) 2 - 5a + \( \frac{3}{2}b \)
(iii) 3x^5 + 4y^3 - 7xy^2 + 3
(iv) \( 2x^2 - \frac{3}{x} + \frac{5}{x^2} + 9 \)
Answer:
(i) The terms of 3 - 7x are 3 and -7x.
(ii) The terms of 2 - 5a + \( \frac{3}{2}b \) are 2, -5a, and \( \frac{3}{2}b \).
(iii) The terms of 3x^5 + 4y^3 - 7xy^2 + 3 are 3x^5, 4y^3, -7xy^2, and 3.
(iv) The terms of \( 2x^2 - \frac{3}{x} + \frac{5}{x^2} + 9 \) are 2x^2, \( -\frac{3}{x} \), \( \frac{5}{x^2} \), and 9.
In simple words: A term is a single part of an expression, separated by + or - signs. List each part as a separate term.

Exam Tip: Count the + and - signs to determine how many terms exist. Each "piece" between operations is one term, including its sign.

 

Question 2. Write down the algebraic expression whose terms are:
(i) 5, -2x
(ii) -3, 5x^2, -7x^5
(iii) a^2, -2b^2, 3c^2, 7
(iv) 2x, \( -\frac{3}{x} \), \( \frac{7y}{x^2} \), 5z, -1
Answer:
(i) The expression built from terms 5 and -2x is 5 - 2x.
(ii) The expression built from terms -3, 5x^2, and -7x^5 is -3 + 5x^2 - 7x^5.
(iii) The expression built from terms a^2, -2b^2, 3c^2, and 7 is a^2 - 2b^2 + 3c^2 + 7.
(iv) The expression built from terms 2x, \( -\frac{3}{x} \), \( \frac{7y}{x^2} \), 5z, and -1 is \( 2x - \frac{3}{x} + \frac{7y}{x^2} + 5z - 1 \).
In simple words: Join all the given terms using + and - signs according to their given signs. A negative term uses a minus sign when written.

Exam Tip: When combining terms, respect the sign that comes with each term. If the term is given as -3, write it with a minus in the final expression.

 

Question 3. State the number of terms in each of the following algebraic expressions:
(i) x ÷ 2 + y - 3
(ii) \( \frac{2x + y - 3}{5} \)
(iii) 5 × ab - 7
(iv) 7 × a + b ÷ 3 - c + 5
Answer:
(i) The terms are x ÷ 2, y, and -3. Hence, the expression has 3 terms.
(ii) The terms are \( \frac{2x}{5} \), \( \frac{y}{5} \), and \( \frac{-3}{5} \). Hence, the expression has 3 terms.
(iii) The terms are 5 × ab and -7. Hence, the expression has 2 terms.
(iv) The terms are 7 × a, b ÷ 3, -c, and 5. Hence, the expression has 4 terms.
In simple words: Count the separate parts. Each part between a + or - sign is one term. Check carefully at fractions too - a numerator with multiple operations means multiple terms.

Exam Tip: When dividing an entire sum by a number, every term in the numerator becomes a separate term in the result. A fraction bar acts like brackets.

 

Question 4. Identify monomials, binomials and trinomials from the following algebraic expressions:
(i) x^2 - 2y^2
(ii) xy + yz + zx
(iii) 5 + 5x
(iv) 5x
(v) \( \frac{5}{x} + 3 \)
(vi) -7
(vii) a^2 - b^2 - c^3 + 5abc
(viii) -5a^2b^2c^2
(ix) 1 + x + x^2 + x^3 + x^4
(x) x ÷ 5 + y ÷ 3
(xi) 5p × ab
(xii) 3 ÷ a - 2 × b + c
Answer: An expression with 1 term is called a monomial; 2 terms is a binomial; 3 terms is a trinomial; and more than 3 terms is a multinomial.
(i) x^2 - 2y^2 has 2 terms → Binomial.
(ii) xy + yz + zx has 3 terms → Trinomial.
(iii) 5 + 5x has 2 terms → Binomial.
(iv) 5x has 1 term → Monomial.
(v) \( \frac{5}{x} + 3 \) has 2 terms → Binomial.
(vi) -7 has 1 term → Monomial.
(vii) a^2 - b^2 - c^3 + 5abc has 4 terms → Multinomial.
(viii) -5a^2b^2c^2 has 1 term → Monomial.
(ix) 1 + x + x^2 + x^3 + x^4 has 5 terms → Multinomial.
(x) x ÷ 5 + y ÷ 3 has 2 terms → Binomial.
(xi) 5p × ab = 5pab has 1 term → Monomial.
(xii) 3 ÷ a - 2 × b + c has 3 terms → Trinomial.
In simple words: Count the number of terms. One term = monomial. Two terms = binomial. Three terms = trinomial. Four or more = multinomial.

Exam Tip: When combining products (like 5p × ab), simplify first to count the terms correctly. A fraction like \( \frac{5}{x} \) is one term, not two.

 

Question 5. Write down the numerical as well as literal coefficient of each of the following monomials:
(i) -7x
(ii) -2x^3y^2
(iii) 6abcd^2
Answer: (i) In the monomial -7x, the numerical coefficient is -7 (the number part) and the literal coefficient is x (the variable part).
(ii) In the monomial -2x^3y^2, the numerical coefficient is -2 and the literal coefficient is x^3y^2.
(iii) In the monomial 6abcd^2, the numerical coefficient is 6 and the literal coefficient is abcd^2.
In simple words: The numerical coefficient is the number at the front. The literal coefficient is all the letters and their powers.

Exam Tip: Always separate the number from the letters. The number multiplying the variables is the numerical coefficient; everything else is the literal coefficient.

 

Question 6. Write the numerical coefficient of each term of the expression: \( x^4 - 5xy^2 + 7x^2 - \frac{2}{3}x - \frac{1}{2} \)
Answer: The individual pieces that make up this expression are \( x^4 \), \( -5xy^2 \), \( 7x^2 \), \( -\frac{2}{3}x \), and \( -\frac{1}{2} \).

For \( x^4 \), the numerical coefficient is 1.

For \( -5xy^2 \), the numerical coefficient is -5.

For \( 7x^2 \), the numerical coefficient is 7.

For \( -\frac{2}{3}x \), the numerical coefficient is \( -\frac{2}{3} \).

For \( -\frac{1}{2} \) (the constant term), the numerical coefficient is \( -\frac{1}{2} \).
In simple words: The number in front of each part of the expression is its numerical coefficient. If you see just a letter with no number, the coefficient is 1. If it is a constant (just a number), that number itself is the coefficient.

Exam Tip: Remember that the numerical coefficient is only the number part - ignore the variables. For a constant term like \( -\frac{1}{2} \), the whole number is the coefficient.

 

Question 7. In \( -7xy^2z^3 \), write down the coefficient of:
(i) x
(ii) 7x
(iii) -xy²
(iv) xy
(v) z³
(vi) -7z
(vii) xyz
(viii) 7yz²
Answer:

The expression is \( -7xy^2z^3 \).

(i) To find the coefficient of x, remove x from the expression, leaving \( -7y^2z^3 \).

(ii) To find the coefficient of 7x, remove 7x, leaving \( -y^2z^3 \).

(iii) To find the coefficient of -xy², remove -xy², leaving \( 7z^3 \).

(iv) To find the coefficient of xy, remove xy, leaving \( -7yz^3 \).

(v) To find the coefficient of z³, remove z³, leaving \( -7xy^2 \).

(vi) To find the coefficient of -7z, remove -7z, leaving \( xy^2z^2 \).

(vii) To find the coefficient of xyz, remove xyz, leaving \( -7yz^2 \).

(viii) To find the coefficient of 7yz², remove 7yz², leaving \( -xyz \).
In simple words: The coefficient of a part is what is left when you take away that part from the full term. Multiply what is left and you get back the original term.

Exam Tip: Always check your answer by multiplying the part you removed by its coefficient - you should get the original term back.

 

Question 8. State true or false:
(i) If 5 is constant and y is variable, then 5y and 5 + y are variables.
(ii) 7x has two terms, 7 and x.
(iii) 5 + xy is a trinomial.
(iv) 7a × bc is a binomial.
(v) 7x³ + 2x² + 3x - 5 is a polynomial.
(vi) 7x³ + 2x² + 3x - 5 is a multinomial.
(vii) \( 2x^2 - \frac{3}{x} \) is a polynomial.
(viii) Coefficient of x in -3xy is -3.
(ix) 3xy, -2yx are like terms.
Answer:

(i) True. Both 5y and 5 + y rely on the value assigned to y to determine their own value, making them variables.

(ii) False. 7x is a single term (monomial). In this term, 7 serves as the numerical coefficient and x is the literal coefficient. They are not two separate terms.

(iii) False. 5 + xy contains only 2 terms, so it is a binomial, not a trinomial.

(iv) False. 7a × bc simplifies to 7abc, which is just one term (monomial).

(v) True. All powers of x in this expression - namely 3, 2, 1, and 0 - are non-negative integers, so it satisfies the definition of a polynomial.

(vi) True. This expression has 4 separate terms, making it a multinomial (a term used for polynomials with more than two terms).

(vii) False. The term \( -\frac{3}{x} \) can be written as \( -3x^{-1} \), which has a negative power of x. Since a polynomial must have only non-negative integer powers, this expression is not a polynomial.

(viii) False. The coefficient of x in -3xy is actually -3y, not just -3, because y is also part of the literal coefficient multiplying x.

(ix) True. Since 3xy and -2yx both have the same literal part (xy, because yx equals xy by the commutative property), they are like terms.
In simple words: Like terms have the exact same letters (variables) raised to the same powers. A polynomial uses only whole number powers of variables. A monomial is one term, a binomial is two terms, and if there are more, it is a multinomial.

Exam Tip: For "true or false" questions on definitions, recall that the defining features matter most - for polynomials, it is non-negative integer powers; for like terms, it is matching variable parts; for naming, it is the count of terms.

 

Question 9. In the following, which pairs contain like terms?
(i) 5x, -2x
(ii) 5x, 5y
(iii) -2xyz, -3x²y²z²
(iv) 10xyz, -10xyz
(v) 2x²y, 2xy²
(vi) 2xy, -3yx
(vii) x, x²
(viii) 6, 6x
(ix) 2xy, 3xyz
Answer:

Two terms are considered like terms when they share the exact same variable part (with the same powers on each variable).

(i) 5x and -2x both have the identical literal part x → Like terms.

(ii) 5x and 5y have different variables (x is not the same as y) → Unlike terms.

(iii) -2xyz and -3x²y²z² have variables with different powers (the first has x¹y¹z¹ while the second has x²y²z²) → Unlike terms.

(iv) 10xyz and -10xyz both have the same literal part xyz → Like terms.

(v) 2x²y and 2xy² have variables with different powers (the first has x² and y¹, the second has x¹ and y²) → Unlike terms.

(vi) 2xy and -3yx both represent the same literal part because xy = yx (multiplication is commutative) → Like terms.

(vii) x and x² have different powers (x¹ vs x²) → Unlike terms.

(viii) 6 contains no variable while 6x contains x → Unlike terms.

(ix) 2xy and 3xyz have different sets of variables (the first has x and y, the second has x, y, and z) → Unlike terms.

Therefore, the pairs containing like terms are (i), (iv), and (vi).
In simple words: Like terms must have exactly the same letters with exactly the same powers. If even one power or one letter is different, they are unlike terms.

Exam Tip: Always write out the powers clearly when comparing terms. Remember that xy and yx are the same, but xy and x²y are different.

 

Question 10. Identify which of the following algebraic expressions are polynomials. If so, write their degrees.
(i) \( \frac{2}{3}x^2 - \frac{5}{7}x - \frac{1}{4} \)
(ii) \( 3x^2 + \frac{2}{x} - 3 \)
(iii) \( x^2 - \frac{2}{3}x^7 - 5x^8 \)
Answer:

An algebraic expression qualifies as a polynomial only when every power of the variable is a non-negative integer (0, 1, 2, 3, ...).

(i) \( \frac{2}{3}x^2 - \frac{5}{7}x - \frac{1}{4} \)

The powers of x are 2, 1, and 0 (in the constant term), all of which are non-negative integers. The highest power is 2. Therefore, this is a polynomial with degree 2.

(ii) \( 3x^2 + \frac{2}{x} - 3 \)

The term \( \frac{2}{x} \) can be rewritten as \( 2x^{-1} \), which has a negative power. Since negative powers are not allowed in polynomials, this is not a polynomial.

(iii) \( x^2 - \frac{2}{3}x^7 - 5x^8 \)

The powers of x are 2, 7, and 8, all non-negative integers. The highest power is 8. Therefore, this is a polynomial with degree 8.
In simple words: A polynomial has only whole number powers (like 1, 2, 3) or zero. If you see a variable in the denominator or a negative power, it is not a polynomial. The degree is just the largest power in the expression.

Exam Tip: When checking if something is a polynomial, look carefully at each term - especially fractions that contain variables. If a variable appears in a denominator, the power is negative and it fails the test.

 

Question 11. Which out of following are expressions with numbers only?
(i) 2y + 3
(ii) (7 × 20) - 8z
(iii) 5 × (21 - 7) + 9 × 2
(iv) 5 - 11n
(v) (5 × 4) - 45 + p
(vi) 3 × (11 + 7) - 24 ÷ 2
Answer:

An expression with numbers only contains no variables - it is made up entirely of constants and operations on those constants.

(i) 2y + 3 has the variable y → Not an expression with numbers only.

(ii) (7 × 20) - 8z has the variable z → Not an expression with numbers only.

(iii) 5 × (21 - 7) + 9 × 2 contains only numbers and arithmetic operations → Expression with numbers only.

(iv) 5 - 11n has the variable n → Not an expression with numbers only.

(v) (5 × 4) - 45 + p has the variable p → Not an expression with numbers only.

(vi) 3 × (11 + 7) - 24 ÷ 2 contains only numbers and arithmetic operations → Expression with numbers only.

Therefore, (iii) and (vi) are expressions with numbers only.
In simple words: If you see any letter (variable) in the expression, it is not a numbers-only expression. If it is just numbers and operations like +, -, ×, ÷, then it is a numbers-only expression.

Exam Tip: Always scan the entire expression for any letters before deciding - even one variable disqualifies it from being a "numbers only" expression.

 

Question 12. Write expression for the following:
(i) 11 added to 2m
(ii) 11 subtracted from 2m
(iii) 3 added to 5 times y
(iv) 3 subtracted from 5 times y
(v) y is multiplied by -8 and then 5 is added to the result
(vi) y is multiplied by 5 and then the result is subtracted from 16
Answer:

(i) 11 added to 2m means 2m + 11 = 2m + 11

(ii) 11 subtracted from 2m means 2m - 11 = 2m - 11

(iii) 3 added to 5 times y means 5y + 3 = 5y + 3

(iv) 3 subtracted from 5 times y means 5y - 3 = 5y - 3

(v) First, y is multiplied by -8 to get -8y. Then 5 is added: -8y + 5 = -8y + 5

(vi) First, y is multiplied by 5 to get 5y. Then this product is subtracted from 16: 16 - 5y = 16 - 5y
In simple words: "Added to" means put a plus sign. "Subtracted from" means the first thing gets the second thing subtracted away from it. "Times" means multiply. Always figure out what you are doing first, then build the expression step by step.

Exam Tip: Pay close attention to the order in "subtracted from" - if it says "A subtracted from B," you write B - A, not A - B.

 

Question 13. Write the following in mathematical form using signs and symbols:
(i) 6 more than thrice a number x
(ii) 7 taken away from y
(iii) 3 less than quotient of x by y
Answer:

(i) Thrice (three times) a number x means 3x. Then "6 more than this" = 3x + 6 = 3x + 6

(ii) "7 taken away from y" means y - 7 = y - 7

(iii) The quotient of x by y (x divided by y) is \( \frac{x}{y} \). Then "3 less than this" means \( \frac{x}{y} - 3 = \frac{x}{y} - 3 \)
In simple words: "Thrice" means three times. "Quotient" means division. "More than" means add, and "less than" or "taken away" means subtract.

Exam Tip: Always identify the key words - "thrice," "quotient," "more than," "less than," "taken away" - and convert them to the correct symbol (×, ÷, +, -).

 

Question 14. Form six expressions using t and 4. Use not more than one number operation and every expression must have t in it.
Answer:

Using a single operation between t and 4, the six expressions that result are:

(i) t + 4

(ii) t - 4

(iii) 4 - t

(iv) 4t (or equivalently, 4 × t)

(v) \( \frac{t}{4} \)

(vi) \( \frac{4}{t} \)
In simple words: Take t and 4 and use one math operation between them: add, subtract, or divide. Multiply counts as an operation too. You can write them in any order as long as each expression has t in it.

Exam Tip: Remember that 4 - t is different from t - 4, and \( \frac{4}{t} \) is different from \( \frac{t}{4} \), so both should be included.

 

Question 15. A student scored x marks in English but the teacher deducted 5 marks for bad handwriting. What was the student's final score in English?
Answer:

The student's score before deduction was x marks.

The teacher took away 5 marks.

The final score = Original score - Deduction = x - 5

Therefore, the student's final score in English is (x - 5) marks.
In simple words: The student earned x marks. Then 5 marks were subtracted. So the answer is x take away 5, which is x - 5.

Exam Tip: Always identify what is being added or removed, and write the expression in the correct order - original amount first, then subtract what is taken away.

 

Question 16. Raju's father's age is 2 years more than 3 times Raju's age. If Raju's present age is y years, then what is his father's age?
Answer:

Raju's current age is y years.

Three times Raju's age is 3y years.

2 years more than this amount is 3y + 2 years.

Therefore, Raju's father's age is (3y + 2) years.
In simple words: Start with Raju's age y. Multiply by 3 to get 3y. Then add 2. The answer is 3y + 2.

Exam Tip: Always follow the order given in the problem - "3 times Raju's age, then 2 more" becomes 3y + 2, not 2 + 3y (though they are the same).

 

Question 17. Mohini is x years old. Express the following in algebraic form:
(i) three times Mohini's age next year.
(ii) four times Mohini's age 3 years ago.
(iii) the present age of Mohini's uncle, if his uncle is 5 times as old as Mohini will be two years from now.
Answer:

Mohini's present age is x years.

(i) Mohini's age one year from now will be (x + 1) years.

Three times her age at that time = 3(x + 1) = 3(x + 1) years

(ii) Mohini's age three years in the past was (x - 3) years.

Four times her age at that time = 4(x - 3) = (4x - 12) years

(iii) Mohini's age two years in the future will be (x + 2) years.

Her uncle's age is 5 times this amount = 5(x + 2) = (5x + 10) years
In simple words: For future ages, add the years to her age. For past ages, subtract the years from her age. Then do the multiplication.

Exam Tip: When a problem mentions "next year" or "years ago," always adjust the age first by adding or subtracting, then apply any multiplication or other operation.

 

Question 18. A bus travels at v km per hour. It is going from Delhi to Jaipur. After the bus has travelled 5 hours, Jaipur is still 20 km away. What is the distance from Delhi to Jaipur?
Answer:

The bus's speed is v km/h (kilometers per hour).

The bus has been traveling for 5 hours.

Distance traveled so far = Speed × Time = v × 5 = 5v km

After 5 hours, the bus still has 20 km left to travel to reach Jaipur.

Total distance from Delhi to Jaipur = Distance already covered + Remaining distance = 5v + 20 km

Therefore, the distance from Delhi to Jaipur is (5v + 20) km.
In simple words: The bus went 5v km in 5 hours. It still needs to go 20 km more. So the total trip is 5v + 20 km.

Exam Tip: In distance-time problems, remember Distance = Speed × Time. Always add up all the parts (what was covered plus what remains) to get the total.

 

Exercise 9.4

 

Question 1. Add:
(a) 2x, 5x
(b) 7p, -2p
(c) 2a + 4b, 9a + 10b
(d) 2xy, 3yx
(e) \( \frac{2}{3}x, 2x \)
Answer:

(a) 2x + 5x = (2 + 5)x = 7x

(b) 7p + (-2p) = (7 - 2)p = 5p

(c) (2a + 4b) + (9a + 10b) = (2a + 9a) + (4b + 10b) = 11a + 14b

(d) 2xy + 3yx = 2xy + 3xy (since yx = xy) = (2 + 3)xy = 5xy

(e) \( \frac{2}{3}x + 2x = \frac{2x + 6x}{3} = \frac{8x}{3} \) = \( \frac{8}{3}x \)
In simple words: When you add terms that have the same variable part, add their numbers (coefficients) together and keep the variable part the same. If the terms have fractions, find a common denominator first.

Exam Tip: Always make sure the variable parts are identical before adding. If they are different (like 2x and 3y), you cannot combine them.

 

Question 2. Subtract:
(a) 12y from 19y
(b) 6z from 5z
(c) 10p from 20p
(d) 13m from 23m
(e) 48n from 50n
Answer:

(a) 19y - 12y = (19 - 12)y = 7y

(b) 5z - 6z = (5 - 6)z = -z

(c) 20p - 10p = (20 - 10)p = 10p

(d) 23m - 13m = (23 - 13)m = 10m

(e) 50n - 48n = (50 - 48)n = 2n
In simple words: When you subtract like terms, subtract their coefficients and keep the variable part. The order matters - the thing you are subtracting from comes first.

Exam Tip: When a problem says "A from B," always compute B - A. This is the most common source of sign errors.

 

Question 3. Solve:
(a) 6x + (9x - 2x)
(b) 11y + 2y + 3x
(c) 11z + 3z + z
(d) 8ab + (-3ab)
(e) 3x - 4x + 7x
(f) 35b - (16b + 9b)
(g) (9a - 3a) + 4a
Answer:

(a) 6x + (9x - 2x) = 6x + 7x = 13x

(b) 11y + 2y + 3x = 13y + 3x (the y terms can be combined, but 3x is separate) = 13y + 3x

(c) 11z + 3z + z = (11 + 3 + 1)z = 15z

(d) 8ab + (-3ab) = (8 - 3)ab = 5ab

(e) 3x - 4x + 7x = (3 - 4 + 7)x = 6x

(f) 35b - (16b + 9b) = 35b - 25b = 10b

(g) (9a - 3a) + 4a = 6a + 4a = 10a
In simple words: Combine terms that have the same variable. Do what is in parentheses first, then combine everything together. Add and subtract the coefficients, keeping the variable the same.

Exam Tip: Always handle parentheses first. When there is a minus sign in front of parentheses, distribute it to change the signs of all terms inside.

 

Question 3. Combine the following:
(b) 11y + 2y + 3x
(c) 11z + 3z + z
(d) 8ab + (−3ab)
(e) 3x − 4x + 7x
(f) 35b − (16b + 9b)
(g) (9a − 3a) + 4a
Answer:
(b) 11y + 2y + 3x = 13y + 3x
(c) 11z + 3z + z = (11 + 3 + 1)z = 15z
(d) 8ab - 3ab = (8 - 3)ab = 5ab
(e) 3x - 4x + 7x = (3 - 4 + 7)x = 6x
(f) 35b - 25b = 10b
(g) 6a + 4a = 10a
In simple words: To combine like terms, add or subtract the numbers in front of the same letters. Letters that are different stay separate.

Exam Tip: Always ensure you combine only like terms - terms with the same variables and exponents. Write the final answer in simplest form.

 

Question 4. Multiply:
(a) 5 × b
(b) x × y
(c) 2a × 3b
(d) a × a
(e) 6a × 2a
Answer:
(a) 5 × b = 5b
(b) x × y = xy
(c) 2a × 3b = (2 × 3) × (a × b) = 6ab
(d) a × a = a²
(e) 6a × 2a = (6 × 2) × (a × a) = 12a²
In simple words: When you multiply numbers by letters, put the numbers and letters together. When the same letter multiplies itself, use an exponent (small number on top) to show how many times it was multiplied.

Exam Tip: Multiply coefficients with coefficients and variables with variables separately, then combine. Use exponent notation when identical variables are multiplied.

 

Question 5. Divide:
(a) 12x ÷ 2x
(b) 24ab ÷ 8a
(c) 9x ÷ 3y
(d) 25a² ÷ 5a
(e) 49xy ÷ 7xy
Answer:
(a) 12x ÷ 2x = \( \frac{12x}{2x} = \frac{12}{2} \times \frac{x}{x} = 6 \)
(b) 24ab ÷ 8a = \( \frac{24ab}{8a} = \frac{24}{8} \times \frac{ab}{a} = 3b \)
(c) 9x ÷ 3y = \( \frac{9x}{3y} = \frac{9}{3} \times \frac{x}{y} = 3\frac{x}{y} \)
(d) 25a² ÷ 5a = \( \frac{25a^2}{5a} = \frac{25}{5} \times \frac{a^2}{a} = 5a \)
(e) 49xy ÷ 7xy = \( \frac{49xy}{7xy} = \frac{49}{7} \times \frac{xy}{xy} = 7 \)
In simple words: To divide, split the numbers and letters separately. Divide the numbers first, then divide the same letters - they cancel out.

Exam Tip: Cancel common factors in numerator and denominator before simplifying. When dividing identical variables, remember that they reduce to 1.

 

Exercise 9.5

 

Question 1. Find the value of:
(i) 3x + 2y when x = 3 and y = 2
(ii) 5x - 3y when x = 2 and y = 5
(iii) a + 2b - 5c when a = 2, b = 3 and c = 1
(iv) 2p + 3q + 4r + pqr when p = 1, q = 2 and r = 3
(v) 3ab + 4bc - 5ca when a = 4, b = 5 and c = 2
Answer:
(i) Putting x = 3 and y = 2 into the expression:
3x + 2y = 3(3) + 2(2) = 9 + 4

\( \therefore 3x + 2y = 13 \)

(ii) Putting x = 2 and y = 5 into the expression:
5x - 3y = 5(2) - 3(5) = 10 - 15

\( \therefore 5x - 3y = -5 \)

(iii) Putting a = 2, b = 3 and c = 1 into the expression:
a + 2b - 5c = 2 + 2(3) - 5(1) = 2 + 6 - 5

\( \therefore a + 2b - 5c = 3 \)

(iv) Putting p = 1, q = 2 and r = 3 into the expression:
2p + 3q + 4r + pqr = 2(1) + 3(2) + 4(3) + (1)(2)(3) = 2 + 6 + 12 + 6

\( \therefore 2p + 3q + 4r + pqr = 26 \)

(v) Putting a = 4, b = 5 and c = 2 into the expression:
3ab + 4bc - 5ca = 3(4)(5) + 4(5)(2) - 5(2)(4) = 60 + 40 - 40

\( \therefore 3ab + 4bc - 5ca = 60 \)
In simple words: Replace each letter with its given number, then calculate step by step following the order of operations.

Exam Tip: Substitute values carefully and show each calculation step. Always follow the correct order of operations (multiply before adding or subtracting).

 

Question 2. Find the value of:
(i) 2x² - 3x + 4 when x = 2
(ii) 4x³ - 5x² - 6x + 7 when x = 3
(iii) 3x³ + 9x² - x + 8 when x = 4
(iv) 2x⁴ - 5x³ + 7x - 3 when x = 2
Answer:
(i) Putting x = 2 into the expression:
2x² - 3x + 4 = 2(2)² - 3(2) + 4 = 2(4) - 6 + 4 = 8 - 6 + 4

\( \therefore 2x^2 - 3x + 4 = 6 \)

(ii) Putting x = 3 into the expression:
4x³ - 5x² - 6x + 7 = 4(3)³ - 5(3)² - 6(3) + 7 = 4(27) - 5(9) - 18 + 7 = 108 - 45 - 18 + 7

\( \therefore 4x^3 - 5x^2 - 6x + 7 = 52 \)

(iii) Putting x = 4 into the expression:
3x³ + 9x² - x + 8 = 3(4)³ + 9(4)² - 4 + 8 = 3(64) + 9(16) - 4 + 8 = 192 + 144 - 4 + 8

\( \therefore 3x^3 + 9x^2 - x + 8 = 340 \)

(iv) Putting x = 2 into the expression:
2x⁴ - 5x³ + 7x - 3 = 2(2)⁴ - 5(2)³ + 7(2) - 3 = 2(16) - 5(8) + 14 - 3 = 32 - 40 + 14 - 3

\( \therefore 2x^4 - 5x^3 + 7x - 3 = 3 \)
In simple words: Work out the powers first, then multiply each term, and finally add or subtract all the results together.

Exam Tip: Calculate exponents before doing multiplications. Show all intermediate steps to avoid arithmetic errors when combining positive and negative values.

 

Question 3. If x = 5, find the values of:
(i) 6 - 7x²
(ii) 3x² + 8x - 10
(iii) 2x³ - 4x² - 6x + 25
Answer:
(i) Putting x = 5 into the expression:
6 - 7x² = 6 - 7(5)² = 6 - 7(25) = 6 - 175

\( \therefore 6 - 7x^2 = -169 \)

(ii) Putting x = 5 into the expression:
3x² + 8x - 10 = 3(5)² + 8(5) - 10 = 3(25) + 40 - 10 = 75 + 40 - 10

\( \therefore 3x^2 + 8x - 10 = 105 \)

(iii) Putting x = 5 into the expression:
2x³ - 4x² - 6x + 25 = 2(5)³ - 4(5)² - 6(5) + 25 = 2(125) - 4(25) - 30 + 25 = 250 - 100 - 30 + 25

\( \therefore 2x^3 - 4x^2 - 6x + 25 = 145 \)
In simple words: Put the number 5 where you see x. Calculate powers and multiplications first, then combine by adding and subtracting from left to right.

Exam Tip: Be careful with negative results - when you subtract a larger number from a smaller one, the answer is negative. Double-check your arithmetic.

 

Question 4. If x = 2, y = 3 and z = 1, find the values of:
(i) x ÷ y
(ii) \( \frac{xy}{z} \)
(iii) \( \frac{2x + 3y - 4z}{3x - z} \)
Answer:
(i) Putting x = 2 and y = 3 into the expression:
x ÷ y = 2 ÷ 3

\( \therefore x \div y = \frac{2}{3} \)

(ii) Putting x = 2, y = 3 and z = 1 into the expression:
\( \frac{xy}{z} = \frac{(2)(3)}{1} = \frac{6}{1} \)

\( \therefore \frac{xy}{z} = 6 \)

(iii) Putting x = 2, y = 3 and z = 1 into the expression:
\[ \implies \frac{2x + 3y - 4z}{3x - z} = \frac{2(2) + 3(3) - 4(1)}{3(2) - 1} = \frac{4 + 9 - 4}{6 - 1} = \frac{9}{5} = 1\frac{4}{5} \]
In simple words: Replace the letters with the numbers given, work out the top and bottom of the fraction separately, then divide to get your answer.

Exam Tip: When working with fractions, simplify the numerator and denominator completely before dividing. Express final answers as improper or mixed fractions as required.

 

Question 5. If a = 2, b = 3 and c = 1, find the value of: a² + b² + c² - 2ab - 2bc - 2ca + 3abc
Answer: Putting a = 2, b = 3 and c = 1 into the expression:
a² + b² + c² - 2ab - 2bc - 2ca + 3abc
= (2)² + (3)² + (1)² - 2(2)(3) - 2(3)(1) - 2(1)(2) + 3(2)(3)(1)
= 4 + 9 + 1 - 12 - 6 - 4 + 18
= 14 - 22 + 18
= -8 + 18

\( \therefore a^2 + b^2 + c^2 - 2ab - 2bc - 2ca + 3abc = 10 \)
In simple words: Replace each letter with its number, calculate all the powers and products, then add the positive terms and subtract the negative terms to get the final result.

Exam Tip: Handle each term separately - calculate the squares first, then the products. Keep track of positive and negative signs throughout the calculation.

 

Question 6. If p = 4, q = -3 and r = 2, find the value of: p³ + q³ - r³ - 3pqr
Answer: Putting p = 4, q = -3 and r = 2 into the expression:
p³ + q³ - r³ - 3pqr
= (4)³ + (-3)³ - (2)³ - 3(4)(-3)(2)
= 64 + (-27) - 8 - (-72)
= 64 - 27 - 8 + 72
= 136 - 35

\( \therefore p^3 + q^3 - r^3 - 3pqr = 101 \)
In simple words: Work out each cube (number times itself three times), then multiply the three numbers together. Pay attention to negative numbers when cubing - a negative times a negative times a negative stays negative.

Exam Tip: Remember that a negative number cubed gives a negative result. When subtracting a negative number, it becomes addition. Always show your sign changes clearly.

 

Question 7. If m = 1, n = 2 and p = 3, find the value of: 2mn⁴ - 15m²n - p
Answer: Putting m = 1, n = 2 and p = 3 into the expression:
2mn⁴ - 15m²n - p
= 2(1)(2)⁴ - 15(1)²(2) - 3
= 2(1)(16) - 15(1)(2) - 3
= 32 - 30 - 3

\( \therefore 2mn^4 - 15m^2n - p = -1 \)
In simple words: Replace the letters, calculate the powers first, then do the multiplications, and finally add or subtract to get your answer.

Exam Tip: Work through powers before multiplication. Write out each substitution step to reduce careless errors when handling multiple terms.

 

Question 8. For x = 5 and y = 2, verify the following:
(i) (x + y)² = x² + 2xy + y²
(ii) (x - y)² = x² - 2xy + y²
(iii) x² - y² = (x + y)(x - y)
(iv) (x + y)² = (x - y)² + 4xy
(v) (x + y)³ = x³ + y³ + 3x²y + 3xy²
Answer:
(i) Putting x = 5 and y = 2 into the identity:
LHS = (x + y)² = (5 + 2)² = (7)² = 49

RHS = x² + 2xy + y² = (5)² + 2(5)(2) + (2)² = 25 + 20 + 4 = 49

Since LHS = RHS, the identity is verified for x = 5 and y = 2.

(ii) Putting x = 5 and y = 2 into the identity:
LHS = (x - y)² = (5 - 2)² = (3)² = 9

RHS = x² - 2xy + y² = (5)² - 2(5)(2) + (2)² = 25 - 20 + 4 = 9

Since LHS = RHS, the identity is verified for x = 5 and y = 2.

(iii) Putting x = 5 and y = 2 into the identity:
LHS = x² - y² = (5)² - (2)² = 25 - 4 = 21

RHS = (x + y)(x - y) = (5 + 2)(5 - 2) = 7 × 3 = 21

Since LHS = RHS, the identity is verified for x = 5 and y = 2.

(iv) Putting x = 5 and y = 2 into the identity:
LHS = (x + y)² = (5 + 2)² = (7)² = 49

RHS = (x - y)² + 4xy = (5 - 2)² + 4(5)(2) = (3)² + 40 = 9 + 40 = 49

Since LHS = RHS, the identity is verified for x = 5 and y = 2.

(v) Putting x = 5 and y = 2 into the identity:
LHS = (x + y)³ = (5 + 2)³ = (7)³ = 343

RHS = x³ + y³ + 3x²y + 3xy² = (5)³ + (2)³ + 3(5)²(2) + 3(5)(2)² = 125 + 8 + 3(25)(2) + 3(5)(4) = 125 + 8 + 150 + 60 = 343

Since LHS = RHS, the identity is verified for x = 5 and y = 2.
In simple words: Replace x and y with the numbers given. Work out the left side and the right side of the equation separately. If both sides give the same answer, the identity is verified.

Exam Tip: Show both sides (LHS and RHS) calculations clearly and state the conclusion explicitly. Verification questions require you to prove both sides are equal, not just find an answer.

 

Exercise 9.6

 

Question 1. State which of the following are equations with a variable. In case of an equation with a variable, identify the variable.
(i) 17 + x = 5
(ii) 2b - 3 = 7
(iii) (y - 7) > 5
(iv) \( \frac{9}{3} = 3 \)
Answer: (i) 17 + x = 5 is an equation with a variable. The variable is x.

(ii) 2b - 3 = 7 is an equation with a variable. The variable is b.

(iii) (y - 7) > 5 is not an equation with a variable because it uses a greater-than symbol (>) instead of an equals sign (=). It is an inequality, not an equation.

(iv) \( \frac{9}{3} = 3 \) is an equation but it does not contain a variable - it is a number sentence showing that 9 divided by 3 equals 3.
In simple words: An equation with a variable is something like "x + 5 = 10" where you have a letter (variable) and an equals sign. If there is no equals sign or no letter, it is not an equation with a variable.

Exam Tip: Remember that an equation must have an equals sign (=). Inequalities use >, <, ≥, or ≤ instead. A statement with only numbers and no variables is a number sentence, not an equation with a variable.

 

Question 1. Which of the following are equations and which are not? Give reasons.
(i) 17 + x = 5
(ii) 2b - 3 = 7
(iii) (y - 7) > 5
(iv) \( \frac{9}{5} = 3 \)
(v) 7 × 3 - 19 = 2
(vi) 5 × 4 - 8 = 3t
(vii) 2p < 15
(viii) 7 = 11 × 5 - 12 × 4
(ix) \( \frac{3}{2}q = 5 \)
Answer: A statement with an equality sign connecting two mathematical expressions is called an equation. An equation must have a variable, and only statements with both an equality sign and at least one variable qualify as equations.

(i) 17 + x = 5 is an equation with variable x.
(ii) 2b - 3 = 7 is an equation with variable b.
(iii) (y - 7) > 5 is not an equation - it is an inequality (uses > instead of =).
(iv) \( \frac{9}{5} = 3 \) is not an equation with a variable - it contains no unknown quantity.
(v) 7 × 3 - 19 = 2 is not an equation with a variable - it is a numerical statement with no unknowns.
(vi) 5 × 4 - 8 = 3t is an equation with variable t.
(vii) 2p < 15 is not an equation - it is an inequality (uses < instead of =).
(viii) 7 = 11 × 5 - 12 × 4 is not an equation with a variable - it is a numerical fact with no unknowns.
(ix) \( \frac{3}{2}q = 5 \) is an equation with variable q.
In simple words: An equation must have an equals sign with a variable on one or both sides. If it has only numbers, or uses < or >, then it is not an equation.

Exam Tip: Look for three key features: an equals sign (not < or >), at least one variable (letter), and two expressions on either side. All three must be present for a statement to be an equation.

 

Question 2. Solve each of the following linear equations:
(i) x + 6 = 8
(ii) x - 2 = -5
(iii) 4x = 6
(iv) \( \frac{x}{2} = 5 \)
(v) 2y - 3 = 2
(vi) 5y + 2 = 4
Answer:
(i) Starting with x + 6 = 8, move the +6 to the other side by subtracting 6 from both sides: x = 8 - 6, so x = 2.

(ii) Starting with x - 2 = -5, move the -2 to the other side by adding 2 to both sides: x = -5 + 2, so x = -3.

(iii) Given 4x = 6, divide both sides by 4 to get x = \( \frac{6}{4} \) = \( \frac{3}{2} \).

(iv) Given \( \frac{x}{2} = 5 \), multiply both sides by 2 to get x = 5 × 2 = 10.

(v) Given 2y - 3 = 2, add 3 to both sides: 2y = 2 + 3 = 5. Now divide both sides by 2: y = \( \frac{5}{2} \).

(vi) Given 5y + 2 = 4, subtract 2 from both sides: 5y = 4 - 2 = 2. Now divide both sides by 5: y = \( \frac{2}{5} \).
In simple words: To solve an equation, move all terms with the variable to one side and numbers to the other. Use opposite operations - if you see +, use -, and if you see ×, use ÷.

Exam Tip: Always show each step of transposing (moving terms to the other side). Write the reason for each step, such as "Transposing +6 to RHS" or "Dividing both sides by 4". Check your answer by substituting back into the original equation.

 

Question 3. Solve the following linear equations:
(i) 5(x + 1) = 25
(ii) 2(3x - 1) = 10
(iii) \( \frac{3x - 1}{4} = 11 \)
Answer:
(i) Start with 5(x + 1) = 25. Divide both sides by 5 to get x + 1 = 5. Move the +1 to the right side to get x = 5 - 1 = 4.

(ii) Start with 2(3x - 1) = 10. Divide both sides by 2 to get 3x - 1 = 5. Add 1 to both sides: 3x = 5 + 1 = 6. Divide both sides by 3: x = \( \frac{6}{3} \) = 2.

(iii) Start with \( \frac{3x - 1}{4} = 11 \). Multiply both sides by 4 to get 3x - 1 = 11 × 4 = 44. Add 1 to both sides: 3x = 44 + 1 = 45. Divide both sides by 3: x = \( \frac{45}{3} \) = 15.
In simple words: When the variable is inside brackets or a fraction, first clear the bracket or fraction by multiplying or dividing both sides. Then move terms and solve as before.

Exam Tip: When brackets are present, divide or multiply to remove them before moving terms. For fractions in the denominator, multiply both sides by that denominator first - this avoids messy fraction arithmetic.

 

Question 4. Solve the following linear equations:
(i) 5x - 6 = 12 - x
(ii) \( \frac{n}{3} + 1 = 4 - n \)
(iii) 5p + 7 = 19 - 2p
(iv) \( 2x + \frac{2}{3} = \frac{5}{2} - x \)
(v) \( \frac{x}{2} - 5 = \frac{x}{3} - 4 \)
(vi) \( 18 - \frac{3y}{4} = 11 + y \)
Answer:
(i) Start with 5x - 6 = 12 - x. Move -x to the left and -6 to the right: 5x + x = 12 + 6, so 6x = 18. Divide both sides by 6: x = 3.

(ii) Start with \( \frac{n}{3} + 1 = 4 - n \). Move all terms with n to one side: \( \frac{n}{3} + n = 4 - 1 = 3 \). Combine on the left: \( \frac{n + 3n}{3} = 3 \), which gives \( \frac{4n}{3} = 3 \). Multiply both sides by 3: 4n = 9. Divide by 4: n = \( \frac{9}{4} \) = 2\( \frac{1}{4} \).

(iii) Start with 5p + 7 = 19 - 2p. Move all p terms to the left and constants to the right: 5p + 2p = 19 - 7, so 7p = 12. Divide by 7: p = \( \frac{12}{7} \) = 1\( \frac{5}{7} \).

(iv) Start with \( 2x + \frac{2}{3} = \frac{5}{2} - x \). Move all x terms to the left: 2x + x = \( \frac{5}{2} - \frac{2}{3} \) = \( \frac{15 - 4}{6} \) = \( \frac{11}{6} \). So 3x = \( \frac{11}{6} \), giving x = \( \frac{11}{18} \).

(v) Start with \( \frac{x}{2} - 5 = \frac{x}{3} - 4 \). Move constant terms to the right and x terms to the left: \( \frac{x}{2} - \frac{x}{3} = -4 + 5 = 1 \). Find common denominator: \( \frac{3x - 2x}{6} = 1 \), so \( \frac{x}{6} = 1 \), giving x = 6.

(vi) Start with \( 18 - \frac{3y}{4} = 11 + y \). Move constants to the left and y terms to the right: 18 - 11 = \( y + \frac{3y}{4} \), so 7 = \( \frac{4y + 3y}{4} \) = \( \frac{7y}{4} \). Multiply both sides by 4: 7y = 28. Divide by 7: y = 4.
In simple words: Gather all variable terms on one side and all numbers on the other side. Combine like terms, then divide to find the value of the variable.

Exam Tip: When variables appear on both sides, move all to one side first before combining. For equations with fractions, find a common denominator early or multiply through by the LCM to clear fractions before solving.

 

Question 1. Fill in the blanks:
(i) In algebra, we use ..... to represent variables (generalized numbers).
(ii) A symbol or letter which can be given various numerical values is called a .....
(iii) If Jaggu's present age is x years, then his age 7 years from now is .....
(iv) If one pen costs Rs.x, then the cost of 9 pens is .....
(v) An equation is a statement that the two expressions are ......
(vi) The numerical coefficient of the monomial -5x²y³ is .....
(vii) 7 less than thrice a number y is .....
(viii) If 3x + 4 = 19, then the value of x is .....
(ix) The number of pencils bought for Rs.x at the rate of Rs.2 per pencil is ......... .
(x) 3p²q and -5pq² are ..... terms.
(xi) The value of the expression 3 - 5x for x = 4 is .....
Answer: (i) letters (literals)
(ii) variable
(iii) (x + 7) years
(iv) Rs.9x
(v) equal
(vi) -5
(vii) 3y - 7
(viii) 5
(ix) \( \frac{x}{2} \)
(x) unlike
(xi) -17
In simple words: These blanks test basic algebra terms. Remember: variables are letters, equations show equality, coefficients are the numbers in front of variables, and substituting a number into an expression means replacing the letter with that number.

Exam Tip: Learn the definitions of key terms: variable, equation, coefficient, like and unlike terms, and expression. Quick mental checks help fill blanks correctly, such as substituting x = 4 into 3 - 5x to verify the answer is -17.

 

Question 2. State whether the following statements are true (T) or false (F):
(i) If x is variable then 5x is also variable.
(ii) If y is variable then y - 5 is also variable.
(iii) The number of angles in a triangle is a variable.
(iv) The value of an algebraic expression changes with the change in the value of the variable.
(v) If the length of a rectangle is twice its breadth, then its area is a constant.
(vi) An equation is satisfied only for a definite value of the variable.
(vii) If x toffees are distributed equally among 5 children, then each child gets 5x toffees.
(viii) t minutes are equal to 60t seconds.
(ix) If x is a negative integer, then -x is a positive integer.
(x) x = 5 is a solution of the equation 3x + 2 = 13.
(xi) 2y - 7 > 13 is an equation.
(xii) 'One third of a number x added to itself gives 8' can be expressed as \( \frac{x}{3} + 8 = x \).
(xiii) The difference between the ages of two sisters Lata and Asha is a variable.
Answer:
(i) True. Since x varies, 5x also varies - its value depends entirely on x.
(ii) True. The expression y - 5 changes whenever y changes, so it is variable.
(iii) False. Every triangle always has exactly 3 angles, which is fixed and constant.
(iv) True. When you change the value of the variable, the result of the expression changes too.
(v) False. If width is b, then length = 2b and area = 2b². Since area depends on b, it is variable, not constant.
(vi) True. A linear equation in one variable has exactly one value that makes it true.
(vii) False. Dividing x toffees among 5 children gives each child \( \frac{x}{5} \) toffees, not 5x.
(viii) True. Since 1 minute = 60 seconds, t minutes = 60t seconds - a direct conversion.
(ix) True. If x is negative, then -x is the opposite, which is positive.
(x) False. Substituting x = 5 gives 3(5) + 2 = 17, not 13.
(xi) False. 2y - 7 > 13 uses the "greater than" symbol, making it an inequality, not an equation.
(xii) False. "One third of a number x added to itself gives 8" means \( \frac{x}{3} + x = 8 \) (or \( x + \frac{x}{3} = 8 \)), not \( \frac{x}{3} + 8 = x \).
(xiii) False. The age difference between two fixed people never changes - if Lata is 5 years older now, she will always be 5 years older, so it is a constant.
In simple words: A quantity is variable if it changes based on something else, and constant if it always stays the same. Equations use =, inequalities use >, <, or ≥, ≤.

Exam Tip: For true/false questions, test with examples - substitute numbers to verify. For statements with variables, remember that the expression itself is variable even if the variable changes. Always read inequality symbols carefully - they are never equations.

 

Question 3. I think of a number x, add 5 to it. The result is then multiplied by 2 and the final result is 24. The correct algebraic statement is
(1) x + 5 × 2 = 24
(2) (x + 5) × 2 = 24
(3) 2 × x + 5 = 24
(4) x + 5 = 2 × 24
Answer: (2) (x + 5) × 2 = 24
In simple words: Add 5 to x to get (x + 5). Then multiply this result by 2, giving 2(x + 5). This final amount equals 24, so the statement is (x + 5) × 2 = 24.

Exam Tip: The order of operations matters. Brackets show which operation happens first - always perform what's inside brackets before multiplying. Option 1 would mean 5 × 2 happens before adding x, which is wrong.

 

Question 4. Which of the following is an equation?
(1) x + 5
(2) 7x
(3) 2y + 3 = 11
(4) 2p < 7
Answer: (3) 2y + 3 = 11
In simple words: An equation must have an equals sign (=) between two expressions. Option 1 and 2 are just expressions with no equals sign. Option 4 has a "less than" symbol, not an equals sign, so it is an inequality.

Exam Tip: The equals sign = is the defining feature of an equation. No equals sign, no equation. Symbols like <, >, ≤, ≥ create inequalities instead.

 

Question 5. If each matchbox contains 48 matchsticks, then the number of matchsticks required to fill n such boxes is
(1) 48 + n
(2) 48 - n
(3) 48 ÷ n
(4) 48n
Answer: (4) 48n
In simple words: If one box has 48 matchsticks, then n boxes have 48 times as many. So you multiply: 48 × n = 48n.

Exam Tip: When you need a total from a group, multiply the amount per item by the number of items. Here, 48 matchsticks per box × n boxes = 48n matchsticks.

 

Question 6. If the perimeter of a regular hexagon is x metres, then the length of each of its sides is
(1) (x + 6) metres
(2) (x - 6) metres
(3) (x ÷ 6) metres
(4) (6 ÷ x) metres
Answer: (3) (x ÷ 6) metres
In simple words: A regular hexagon has 6 equal sides. If the perimeter (total distance around) is x, then each side is \( \frac{x}{6} \) metres, which is the same as x ÷ 6.

Exam Tip: For regular (all sides equal) polygons, divide the perimeter by the number of sides. A regular hexagon has 6 sides, so divide by 6. A square has 4 sides, so divide by 4, and so on.

 

Question 7. x = 3 is the solution of the equation
(1) x + 7 = 4
(2) x + 10 = 7
(3) x + 7 = 10
(4) x + 3 = 7
Answer: (3) x + 7 = 10
In simple words: Test x = 3 in each equation. In option 3: 3 + 7 = 10, which is true. In all other options, the equation becomes false when x = 3.

Exam Tip: To check if a value is a solution, substitute it into the equation and see if both sides match. If they do, it is a solution. If not, try another option.

 

Question 7. Which equation is satisfied when x = 3?
(i) x + 7 = 4
(ii) x + 10 = 7
(iii) x + 7 = 10
(iv) x + 3 = 7
Answer: Put x = 3 into each equation one at a time. For (i): 3 + 7 = 10, which is not 4. For (ii): 3 + 10 = 13, which is not 7. For (iii): 3 + 7 = 10, which matches the equation. For (iv): 3 + 3 = 6, which is not 7. Since option (iii) works, that is the solution.
In simple words: Try putting 3 in place of x in each equation. The one where both sides match is the right answer.

Exam Tip: Always substitute the given value carefully into each equation and check if left side equals right side - this method works for all such questions.

 

Question 8. The solution of the equation 3x - 2 = 10 is
(a) x = 1
(b) x = 2
(c) x = 3
(d) x = 4
Answer: (d) x = 4
Start with 3x - 2 = 10. Add 2 to both sides to get 3x = 12. Divide both sides by 3 to find x = 4.
In simple words: Move the -2 to the other side by adding. Then divide 12 by 3 to get the answer.

Exam Tip: When solving equations, do the same operation on both sides - this keeps the equation balanced and leads to the correct solution.

 

Question 9. The operation not involved in forming the expression \( 5x + \frac{5}{x} \) from the variable x and number 5 is
(a) addition
(b) subtraction
(c) multiplication
(d) division
Answer: (b) subtraction
In the expression \( 5x + \frac{5}{x} \): the term \( \frac{5}{x} \) uses division, the term 5x uses multiplication, and the plus sign uses addition. There is no subtraction in this expression.
In simple words: Look at each part of the expression. Division is in \( \frac{5}{x} \), multiplication is in 5x, addition is the plus sign. Subtraction does not appear.

Exam Tip: Read the expression carefully from left to right and identify which operations (addition, subtraction, multiplication, division) actually show up.

 

Question 10. The quotient of x by 3 added to 7 is written as
(a) \( \frac{x}{3} + 7 \)
(b) \( \frac{3}{x} + 7 \)
(c) \( \frac{x + 3}{7} \)
(d) \( \frac{x}{3 \times 7} \)
Answer: (a) \( \frac{x}{3} + 7 \)
The quotient of x by 3 means x divided by 3, which is \( \frac{x}{3} \). When we add 7 to this quotient, we get \( \frac{x}{3} + 7 \).
In simple words: "Quotient" means divide. So quotient of x by 3 is x divided by 3. Then add 7 to that.

Exam Tip: Remember that "quotient of A by B" means A ÷ B or \( \frac{A}{B} \) - quotient always means division, with the first number on top and the second on the bottom.

 

Question 11. If there are x chairs in a row, then the number of persons that can be seated in 8 rows are
(a) 64
(b) x + 8
(c) 8x
(d) none of these
Answer: (c) 8x
If each row has x chairs, then 8 rows will have 8 times as many chairs. This equals 8 × x = 8x.
In simple words: One row has x chairs. Eight rows have 8 times more chairs, so that is 8x chairs total.

Exam Tip: When multiplying a number by a variable, write the number first followed by the variable (8x, not x8) - this is the standard way to write algebraic terms.

 

Question 12. If Arshad earns Rs x per day and spends Rs y per day, then his saving for the month of March is
(a) Rs (31x - y)
(b) Rs 31(x - y)
(c) Rs 31(x + y)
(d) Rs 31(y - x)
Answer: (b) Rs 31(x - y)
Each day, Arshad saves (earnings - spending) = Rs (x - y). March has 31 days, so his total savings is 31 × (x - y) = Rs 31(x - y).
In simple words: He saves (x - y) every day. Multiply this by 31 days to find his savings for the whole month.

Exam Tip: For time-based problems, first find the daily amount, then multiply by the number of days - this structure works for many real-world algebra questions.

 

Question 13. If the length of a rectangle is 3 times its breadth and the breadth is x units, then its perimeter is
(a) 4x units
(b) 6x units
(c) 8x units
(d) 10x units
Answer: (c) 8x units
Breadth = x units. Length = 3x units. Perimeter = 2(length + breadth) = 2(3x + x) = 2(4x) = 8x units.
In simple words: Add length and breadth to get 3x + x = 4x. Then multiply by 2 to get the full perimeter.

Exam Tip: Always use the correct perimeter formula: Perimeter = 2(length + breadth) for rectangles - forgetting to multiply by 2 is a common mistake.

 

Question 14. Rashmi has a sum of Rs x. She spent Rs 800 on grocery, Rs 600 on clothes and Rs 500 on education and received Rs 200 as a gift. How much money (in Rs) is left with her?
(a) x - 1700
(b) x - 1900
(c) x + 200
(d) x - 2100
Answer: (a) x - 1700
Total spending = 800 + 600 + 500 = Rs 1900. After spending, she has x - 1900. She then gets Rs 200 as a gift. Final amount = (x - 1900) + 200 = x - 1700.
In simple words: Add up all the money she spent. Subtract that from x. Then add the gift she received.

Exam Tip: Break money problems into steps: find total spent, subtract from what she had, then add any money received - this prevents calculation errors.

 

Question 15. For any two integers a and b, which of the following suggests that the operation of addition is commutative?
(a) a × b = b × a
(b) a + b = b + a
(c) a - b = b - a
(d) a + b > a
Answer: (b) a + b = b + a
The commutative property of addition says that when you add two numbers, the order does not matter. The result is the same whether you add a to b or b to a.
In simple words: Addition is commutative because 2 + 3 gives the same answer as 3 + 2 - both equal 5.

Exam Tip: Remember "commutative" means you can swap the order of the numbers and get the same result - addition and multiplication are commutative, but subtraction and division are not.

 

Question 16. The number of unlike terms in the expression \( 2x^2y - 3y^2z - 5yx^2 + yz(x + y) + 7 \) is
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (b) 4
Expand and simplify: \( 2x^2y - 3y^2z - 5yx^2 + yz(x + y) + 7 = 2x^2y - 5yx^2 - 3y^2z + yzx + y^2z + 7 = -3x^2y - 2y^2z + xyz + 7 \). The four unlike terms are: \( -3x^2y, -2y^2z, xyz \), and 7.
In simple words: First expand the brackets. Then group the like terms together. Count how many different groups you have.

Exam Tip: Unlike terms have different variable parts - always expand brackets and combine like terms before counting unlike terms.

 

Question 17. The value of the polynomial \( 3x^2 - 5x + 3 \) when x = 1 is
(a) -1
(b) 0
(c) 1
(d) 11
Answer: (c) 1
Replace x with 1: \( 3(1)^2 - 5(1) + 3 = 3 - 5 + 3 = 1 \).
In simple words: Put 1 everywhere you see x. Then calculate step by step from left to right.

Exam Tip: When substituting into a polynomial, replace every x carefully, then follow order of operations (powers first, then multiplication and division, then addition and subtraction).

 

Statement I-II Type Questions

 

Question 18. Statement I: The number of terms in the expression x + 2y + y is 3. Statement II: 3x + 4 = 0 is a linear equation.
(a) Statement I is true but statement II is false.
(b) Statement I is false but statement II is true.
(c) Both Statement I and statement II are true.
(d) Both Statement I and statement II are false.
Answer: (b) Statement I is false but statement II is true
Statement I: Simplify x + 2y + y to x + 3y. This has 2 terms (x and 3y), not 3. Statement I is false. Statement II: The equation 3x + 4 = 0 has x with the highest power 1, which makes it a linear equation. Statement II is true.
In simple words: Combine like terms in Statement I - you get 2 terms, not 3. In Statement II, a linear equation has x to the first power only.

Exam Tip: Always simplify expressions by combining like terms before counting how many terms remain - this is where most mistakes happen.

 

Question 19. Statement I: If John's present age is x years, then five less than double his age is (2x - 5) years. Statement II: 3x × 4y is an algebraic expression with only one term.
(a) Statement I is true but statement II is false.
(b) Statement I is false but statement II is true.
(c) Both Statement I and statement II are true.
(d) Both Statement I and statement II are false.
Answer: (c) Both Statement I and statement II are true
Statement I: Double John's age = 2x. Five less than double his age = 2x - 5. Statement I is true. Statement II: Multiply 3x × 4y = 12xy, which is a single term. Statement II is true.
In simple words: In Statement I, "double" means multiply by 2, then subtract 5. In Statement II, multiplying the two terms gives one term.

Exam Tip: When two terms are multiplied together, the result is one term (like 3x × 4y = 12xy) - but when they are added or subtracted, they stay separate.

 

Question 20. Statement I: The value of the expression 4y + 3x when x = 2 and y = 3 is 22. Statement II: An expression is a combination of terms connected by mathematical operations of addition, subtraction or both.
(a) Statement I is true but statement II is false.
(b) Statement I is false but statement II is true.
(c) Both Statement I and statement II are true.
(d) Both Statement I and statement II are false.
Answer: (b) Statement I is false but statement II is true
Statement I: Substitute x = 2 and y = 3: 4(3) + 3(2) = 12 + 6 = 18, not 22. Statement I is false. Statement II: An expression is indeed made by combining terms using addition, subtraction, or both. Statement II is true.
In simple words: In Statement I, calculate each part carefully: 4 times 3 is 12, 3 times 2 is 6, and 12 + 6 equals 18. In Statement II, think of how expressions like 2x + 3y or 5a - b are built.

Exam Tip: When substituting values, write out each step - this prevents careless arithmetic mistakes and makes it easy to check your work.

 

Check Your Progress

 

Question 1. Look at the following matchstick pattern of polygons. Complete the table. Also write the general rule that gives the number of matchsticks.
Answer: Each polygon (house-shape) is made with 5 matchsticks. When two adjacent polygons are joined, they share 1 matchstick. So for every new polygon added after the first one, you need 4 more matchsticks.

Looking at the pattern: 5 = 4(1) + 1, 9 = 4(2) + 1, 13 = 4(3) + 1, 17 = 4(4) + 1, 21 = 4(5) + 1. The general rule is 4n + 1, where n is the number of polygons.
In simple words: Multiply the number of polygons by 4, then add 1. This formula always gives the right number of matchsticks.

Exam Tip: When finding a general rule from a table, look for the pattern: find what changes each time, then write a formula using n for the position or count.

 

Question 2. Consider the expression \( \frac{3}{2}x^2y - \frac{1}{2}xy^2 + 6x^2y^2 \). (i) How many terms are there? What do you call such an expression? (ii) List out the terms. (iii) In the term \( -\frac{1}{2}xy^2 \), write down the numerical coefficient and the literal coefficient. (iv) In the term \( -\frac{1}{2}xy^2 \), what is the coefficient of x?
Answer:
(i) The given expression \( \frac{3}{2}x^2y - \frac{1}{2}xy^2 + 6x^2y^2 \) has 3 terms. Since it has three terms, it is called a trinomial.

(ii) The three terms are: \( \frac{3}{2}x^2y \), \( -\frac{1}{2}xy^2 \), and \( 6x^2y^2 \)

(iii) In the term \( -\frac{1}{2}xy^2 \):
- Numerical coefficient = \( -\frac{1}{2} \)
- Literal coefficient = \( xy^2 \)

(iv) In the term \( -\frac{1}{2}xy^2 \), to find the coefficient of x, remove x from the term: \( -\frac{1}{2}xy^2 = -\frac{1}{2} \times x \times y^2 \). The coefficient of x is \( -\frac{1}{2}y^2 \).
In simple words: Count the terms in the expression - you get 3, so it is a trinomial. The numerical coefficient is just the number in front. The literal coefficient is the variable part. To find the coefficient of one variable, see what is left when you remove that variable.

Exam Tip: Always separate the number part (numerical coefficient) from the variable part (literal coefficient) - this makes identifying coefficients of specific variables much easier.

 

Question 3. Write an algebraic expression for each of the following: (i) If 1 metre cloth costs Rs x, then what is cost of 6 metre cloth? (ii) If the cost of a notebook is Rs x and the cost of a book is Rs y, then what is the cost of 5 notebooks and 2 books? (iii) The score of Ragini in Mathematics is 23 more than two-third of her score in English. If she scores x marks in English, what is her score in Mathematics? (iv) If the length of a side of a regular pentagon is x cm, then what is the perimeter of the pentagon?
Answer:
(i) Cost of 1 metre = Rs x. Therefore, cost of 6 metre = Rs (6 × x) = Rs 6x

(ii) Cost of 1 notebook = Rs x, so cost of 5 notebooks = Rs 5x. Cost of 1 book = Rs y, so cost of 2 books = Rs 2y. Total cost = Rs (5x + 2y)

(iii) Ragini's English score = x marks. Two-third of her English score = \( \frac{2}{3}x \). Mathematics score = 23 more than two-third of English score = \( \frac{2}{3}x + 23 \)

(iv) A regular pentagon has 5 equal sides. If each side is x cm, the perimeter = 5x cm
In simple words: For each problem, identify what you know, then write it as a formula using the variables given. Multiply when you have multiple items of the same kind. Add when combining different things.

Exam Tip: Read word problems carefully - "cost of 6 metres" means multiply by 6, "more than" means add, and "two-thirds of" means multiply by 2/3. Watch for these signal words.

 

Question 4. When x = 4 and y = 2, find the value of:
(i) x + y
(ii) x - y
(iii) x² + 2
(iv) x² - 2xy + y²
Answer: Given x = 4 and y = 2.
(i) Putting x = 4 and y = 2 into x + y:
\( x + y = 4 + 2 = 6 \)
(ii) Putting x = 4 and y = 2 into x - y:
\( x - y = 4 - 2 = 2 \)
(iii) Putting x = 4 into x² + 2:
\( x^2 + 2 = (4)^2 + 2 = 16 + 2 = 18 \)
(iv) Putting x = 4 and y = 2 into x² - 2xy + y²:
\( x^2 - 2xy + y^2 = (4)^2 - 2(4)(2) + (2)^2 = 16 - 16 + 4 = 4 \)
In simple words: Replace each letter with its number value. Then work out the math step by step, following the order of operations (squares first, then multiply, then add or subtract).

Exam Tip: Always write out the substitution clearly and show each step of the working - careless arithmetic mistakes are easy to make when you skip steps.

 

Question 5. When a = 3, b = 0 and c = 4, find the value of:
(i) ab + 2bc + 3ca + 4abc
(ii) a³ + b³ + c³ - 3abc
Answer: Given a = 3, b = 0 and c = 4.
(i) Putting a = 3, b = 0 and c = 4 into ab + 2bc + 3ca + 4abc:
\( ab + 2bc + 3ca + 4abc = (3)(0) + 2(0)(4) + 3(4)(3) + 4(3)(0)(4) = 0 + 0 + 36 + 0 = 36 \)
(ii) Putting a = 3, b = 0 and c = 4 into a³ + b³ + c³ - 3abc:
\( a^3 + b^3 + c^3 - 3abc = (3)^3 + (0)^3 + (4)^3 - 3(3)(0)(4) = 27 + 0 + 64 - 0 = 91 \)
In simple words: Write the number value in place of each letter, then multiply and add following the order of operations. When a variable has the value 0, any product containing it becomes 0.

Exam Tip: Notice how b = 0 makes several terms disappear - always spot these shortcuts early to save time and reduce arithmetic errors.

 

Question 6. Solve the following linear equations:
(i) \( 2x - 1\frac{1}{2} = 4\frac{1}{2} \)
(ii) 3(y - 1) = 2(y + 1)
(iii) n - 3 = 5n - 5
(iv) \( \frac{1}{3}(7x - 1) = \frac{1}{4} \)
Answer:
(i) We have:
\( 2x - 1\frac{1}{2} = 4\frac{1}{2} \)
\( \Rightarrow 2x - \frac{3}{2} = \frac{9}{2} \)
\( \Rightarrow 2x = \frac{9}{2} + \frac{3}{2} \) [Moving \( -\frac{3}{2} \) to the right side]
\( \Rightarrow 2x = \frac{12}{2} \)
\( \Rightarrow 2x = 6 \)
\( \Rightarrow x = 3 \)
Therefore x = 3.
(ii) We have:
\( 3(y - 1) = 2(y + 1) \)
\( \Rightarrow 3y - 3 = 2y + 2 \) [Removing brackets]
\( \Rightarrow 3y - 2y = 2 + 3 \) [Moving -3 to the right side and 2y to the left side]
\( \Rightarrow y = 5 \)
Therefore y = 5.
(iii) We have:
\( n - 3 = 5n - 5 \)
\( \Rightarrow n - 5n = -5 + 3 \) [Moving -3 to the right side and 5n to the left side]
\( \Rightarrow -4n = -2 \)
\( \Rightarrow n = \frac{-2}{-4} \)
\( \Rightarrow n = \frac{1}{2} \)
Therefore \( n = \frac{1}{2} \).
(iv) We have:
\( \frac{1}{3}(7x - 1) = \frac{1}{4} \)
\( \Rightarrow 7x - 1 = \frac{1}{4} \times 3 \) [Multiplying both sides by 3]
\( \Rightarrow 7x - 1 = \frac{3}{4} \)
\( \Rightarrow 7x = \frac{3}{4} + 1 \) [Moving -1 to the right side]
\( \Rightarrow 7x = \frac{3 + 4}{4} \)
\( \Rightarrow 7x = \frac{7}{4} \)
\( \Rightarrow x = \frac{7}{4} \times \frac{1}{7} \)
Therefore \( x = \frac{1}{4} \).
In simple words: To solve, move all terms with the variable to one side and all numbers to the other side. Use the opposite operation (add becomes subtract, multiply becomes divide) when you move a term across the equals sign.

Exam Tip: Always show the transposition step and the operation used - examiners award marks for method even if arithmetic slips occur. Check your answer by substituting it back into the original equation.