Selina Concise Solutions for ICSE Class 7 Mathematics Chapter 12 Simple Linear Equations Including Word Problems

Simple Linear Equations (Including Word Problems)

Points to Remember

Equation: This is a mathematical statement showing that two expressions are equal to each other.

Solving an Equation: Solving an equation means finding the value of the unknown variable in it. An equation does not change if:

• The same number is added to both sides of the equation.
• The same number is subtracted from both sides of the equation.
• Both sides are multiplied by the same number.
• Both sides are divided by the same non-zero number.
• When moving a term from one side to another, its sign changes (transposition) - positive becomes negative and negative becomes positive, while multiplication becomes division and division becomes multiplication.

Inequation: An inequation is a statement showing inequality between two mathematical expressions with a single variable where the highest power is one.

Replacement Set: The set of values from which we can choose the value of the variable for an inequation.

Solution Set: The group of values from the replacement set that make the inequation true.

Properties of Inequations: Adding, subtracting, multiplying, or dividing both sides by the same positive number keeps the inequality sign the same. However, multiplying or dividing by a negative number flips the inequality sign.

Exercise 12(A)

Solve the following equations:

Question 1. x + 5 = 10
Answer:
\( x + 5 = 10 \)
\( \implies x = 10 - 5 \)
\( \implies x = 5 \)
In simple words: Move 5 to the other side of the equals sign. When it moves, its sign changes from plus to minus. So, subtract 5 from 10 to get 5.

Exam Tip: Remember to always change the sign of a term when shifting it to the other side of the equals sign.

Question 2. 2 + y = 7
Answer:
\( 2 + y = 7 \)
\( \implies y = 7 - 2 \)
\( \implies y = 5 \)
In simple words: Shift 2 to the right side of the equals sign. Since it was positive, it becomes negative. Subtracting 2 from 7 gives 5.

Exam Tip: Check your answer by putting the value back into the equation: 2 + 5 = 7, which is correct.

Question 3. a - 2 = 6
Answer:
\( a - 2 = 6 \)
\( \implies a = 6 + 2 \)
\( \implies a = 8 \)
In simple words: When you move the minus 2 to the other side, it turns into plus 2. Adding 2 to 6 gives 8.

Exam Tip: A minus term on one side always becomes a plus term when moved to the other side.

Question 4. x - 5 = 8
Answer:
\( x - 5 = 8 \)
\( \implies x = 8 + 5 \)
\( \implies x = 13 \)
In simple words: Take -5 to the right side, which changes it to +5. Add 5 to 8 to find that the variable is 13.

Exam Tip: Be careful with signs. Changing the sign of the constant term correctly is key to solving basic equations.

Question 5. 5 - d = 12
Answer:
\( 5 - d = 12 \)
\( \implies -d = 12 - 5 \)
\( \implies -d = 7 \)
\( \implies d = -7 \)
In simple words: Move 5 to the right side to get -d equals 7. To find positive d, change the sign of both sides, giving d equals -7.

Exam Tip: Do not forget the minus sign attached to the variable; you must multiply or divide by -1 at the end to get the positive variable.

Question 6. 3p = 12
Answer:
\( 3p = 12 \)
\( \implies p = \frac{12}{3} \)
\( \implies p = 4 \)
In simple words: Since 3 is multiplied by p, divide 12 by 3 when you move 3 to the other side. This gives 4.

Exam Tip: A coefficient multiplied by a variable becomes a divisor on the other side of the equation.

Question 7. 14 = 7m
Answer:
\( 14 = 7m \)
\( \implies m = \frac{14}{7} \)
\( \implies m = 2 \)
In simple words: To find m, divide 14 by 7. This gives a final value of 2.

Exam Tip: Keep the variable on either side as long as you divide correctly. 14 = 7m is the same as 7m = 14.

Question 8. 2x = 0
Answer:
\( 2x = 0 \)
\( \implies x = \frac{0}{2} \)
\( \implies x = 0 \)
In simple words: Divide 0 by 2. Any number divided into 0 is still 0, so x is 0.

Exam Tip: Dividing zero by any non-zero number always results in zero.

Question 9. \( \frac{x}{9} = 2 \)
Answer:
\( \frac{x}{9} = 2 \)
\( \implies x = 2 \times 9 \)
\( \implies x = 18 \)
In simple words: Since x is divided by 9, multiply the other side by 9 to solve it. 2 times 9 is 18.

Exam Tip: A number in the denominator moves to the other side as a multiplier.

Question 10. \( \frac{y}{-12} = -4 \)
Answer:
\( \frac{y}{-12} = -4 \)
\( \implies y = (-4) \times (-12) \)
\( \implies y = 48 \)
In simple words: Multiply -4 by -12. Multiplying two negative numbers always gives a positive result, which is 48.

Exam Tip: Remember the sign rule: minus times minus becomes plus.

Question 11. 8x - 2 = 38
Answer:
\( 8x - 2 = 38 \)
\( \implies 8x = 38 + 2 \)
\( \implies 8x = 40 \)
\( \implies x = \frac{40}{8} \)
\( \implies x = 5 \)
In simple words: First, add 2 to 38 to get 40. Then, divide 40 by 8 to find that x is 5.

Exam Tip: Perform addition or subtraction first before doing division to isolate the variable.

Question 12. 2x + 5 = 5
Answer:
\( 2x + 5 = 5 \)
\( \implies 2x = 5 - 5 \)
\( \implies 2x = 0 \)
\( \implies x = \frac{0}{2} \)
\( \implies x = 0 \)
In simple words: Subtract 5 from 5, leaving 0 on the right side. Dividing 0 by 2 gives 0.

Exam Tip: Do not get confused when the right side becomes zero; the variable can still have a value of zero.

Question 13. 5x - 1 = 74
Answer:
\( 5x - 1 = 74 \)
\( \implies 5x = 74 + 1 \)
\( \implies 5x = 75 \)
\( \implies x = \frac{75}{5} \)
\( \implies x = 15 \)
In simple words: Add 1 to 74 to get 75. Then, divide 75 by 5, which gives 15.

Exam Tip: Always do operations step-by-step: first transpose terms without the variable, then divide by the coefficient.

Question 14. 14 = 27 - x
Answer:
\( 14 = 27 - x \)
\( \implies x = 27 - 14 \)
\( \implies x = 13 \)
In simple words: Shift -x to the left side to make it positive, and move 14 to the right side to subtract it from 27. This gives 13.

Exam Tip: Moving the negative variable to the other side is a quick way to make it positive.

Question 15. 10 + 6a = 40
Answer:
\( 10 + 6a = 40 \)
\( \implies 6a = 40 - 10 \)
\( \implies 6a = 30 \)
\( \implies a = \frac{30}{6} \)
\( \implies a = 5 \)
In simple words: Subtract 10 from 40 to get 30. Then divide 30 by 6 to find that a is 5.

Exam Tip: Ensure that you subtract the correct constant term before dividing.

Question 16. \( c - \frac{1}{2} = \frac{1}{3} \)
Answer:
\( c - \frac{1}{2} = \frac{1}{3} \)
\( \implies c = \frac{1}{3} + \frac{1}{2} \)
\( \implies c = \frac{2 + 3}{6} \)
\( \implies c = \frac{5}{6} \)
In simple words: Move half to the right side and add it to one-third. Find the common denominator of 6 to add them, which gives five-sixths.

Exam Tip: Always find the least common multiple (LCM) of denominators when adding fractions.

Question 17. \( \frac{a}{15} - 2 = 0 \)
Answer:
\( \frac{a}{15} - 2 = 0 \)
\( \implies \frac{a}{15} = 2 \)
\( \implies a = 2 \times 15 \)
\( \implies a = 30 \)
In simple words: Move the minus 2 to the right side to get 2. Then multiply 2 by 15, which gives 30.

Exam Tip: Isolate the fraction term first before multiplying by the denominator.

Question 18. 12 = c - 2
Answer:
\( 12 = c - 2 \)
\( \implies 12 + 2 = c \)
\( \implies 14 = c \)
\( \implies c = 14 \)
In simple words: Add 2 to 12. This gives the value of c, which is 14.

Exam Tip: The position of the variable on the left or right does not change how you isolate it.

Question 19. 4 = x - 2.5
Answer:
\( 4 = x - 2.5 \)
\( \implies 4 + 2.5 = x \)
\( \implies 6.5 = x \)
\( \implies x = 6.5 \)
In simple words: Move -2.5 to the left side and add it to 4 to find that x is 6.5.

Exam Tip: Take care when adding decimal numbers to whole numbers by aligning decimal places.

Question 20. \( y + 5 = 8\frac{1}{4} \)
Answer:
\( y + 5 = 8\frac{1}{4} \)
\( \implies y + 5 = \frac{33}{4} \)
\( \implies y = \frac{33}{4} - 5 \)
\( \implies y = \frac{33 - 20}{4} \)
\( \implies y = \frac{13}{4} = 3\frac{1}{4} \)
In simple words: Change the mixed number to an improper fraction, which is thirty-three fourths. Then subtract 5 by using a common denominator, giving thirteen-fourths, or three and one-quarter.

Exam Tip: Convert mixed fractions into improper fractions before performing subtraction or addition.

Question 21. \( x + \frac{1}{4} = -\frac{3}{8} \)
Answer:
\( x + \frac{1}{4} = -\frac{3}{8} \)
\( \implies x = -\frac{3}{8} - \frac{1}{4} \)
\( \implies x = \frac{-3 - 2}{8} \)
\( \implies x = -\frac{5}{8} \)
In simple words: Subtract one-fourth from minus three-eighths. Use the common denominator of 8, which gives minus five-eighths.

Exam Tip: Be cautious when subtracting from a negative fraction, as the result will become more negative.

Question 22. p + 0.02 = 0.08
Answer:
\( p + 0.02 = 0.08 \)
\( \implies p = 0.08 - 0.02 \)
\( \implies p = 0.06 \)
In simple words: Subtract 0.02 from 0.08 to find the value of p, which is 0.06.

Exam Tip: Align the decimal points carefully when subtracting decimals to avoid place value errors.

Question 23. \( p - 12 = 2\frac{2}{3} \)
Answer:
\( p - 12 = 2\frac{2}{3} \)
\( \implies p - 12 = \frac{8}{3} \)
\( \implies p = \frac{8}{3} + 12 \)
\( \implies p = \frac{8 + 36}{3} \)
\( \implies p = \frac{44}{3} = 14\frac{2}{3} \)
In simple words: Convert the mixed fraction to eight-thirds, then add 12 to it. Using a common denominator of 3 gives forty-four thirds, which simplifies to fourteen and two-thirds.

Exam Tip: Express your final fraction in mixed form if the question uses mixed fractions.

Question 24. -3x = 15
Answer:
\( -3x = 15 \)
\( \implies x = \frac{15}{-3} \)
\( \implies x = -5 \)
In simple words: Divide 15 by -3 to isolate x. A positive divided by a negative gives a negative result, which is -5.

Exam Tip: Remember that a positive number divided by a negative number is negative.

Question 25. 1.3h = 39
Answer:
\( 1.3h = 39 \)
\( \implies h = \frac{39}{1.3} \)
\( \implies h = \frac{39 \times 10}{13} \)
\( \implies h = 30 \)
In simple words: Divide 39 by 1.3. Multiply the top and bottom by 10 to make it thirty-nine times ten divided by thirteen, which gives 30.

Exam Tip: Multiply both numerator and denominator by 10 to clear decimals in fractions easily.

Question 26. \( \frac{5}{8}n = 20 \)
Answer:
\( \frac{5}{8}n = 20 \)
\( \implies 5n = 20 \times 8 \)
\( \implies 5n = 160 \)
\( \implies n = \frac{160}{5} \)
\( \implies n = 32 \)
In simple words: Multiply 20 by 8 to get 160. Then, divide 160 by 5 to find that n is 32.

Exam Tip: You can also solve this in one step by multiplying both sides by the reciprocal, \( \frac{8}{5} \).

Question 27. \( \frac{3}{16}m = 21 \)
Answer:
\( \frac{3}{16}m = 21 \)
\( \implies 3m = 21 \times 16 \)
\( \implies 3m = 336 \)
\( \implies m = \frac{336}{3} \)
\( \implies m = 112 \)
In simple words: Multiply 21 by 16 to get 336. Then divide 336 by 3 to find that m is 112.

Exam Tip: Simplify the calculation by dividing first: \( m = \frac{21 \times 16}{3} = 7 \times 16 = 112 \).

Question 28. 2a - 3 = 5
Answer:
\( 2a - 3 = 5 \)
\( \implies 2a = 5 + 3 \)
\( \implies 2a = 8 \)
\( \implies a = \frac{8}{2} \)
\( \implies a = 4 \)
In simple words: Add 3 to 5 to get 8. Then, divide 8 by 2 to find that a is 4.

Exam Tip: Move the constant first, then divide by the coefficient of the variable.

Question 29. 3p - 1 = 8
Answer:
\( 3p - 1 = 8 \)
\( \implies 3p = 8 + 1 \)
\( \implies 3p = 9 \)
\( \implies p = \frac{9}{3} \)
\( \implies p = 3 \)
In simple words: Add 1 to 8 to make 9. Then, divide 9 by 3 to show that p is 3.

Exam Tip: Always perform transposition first before isolating the variable through division.

Question 30. 9y - 7 = 20
Answer:
\( 9y - 7 = 20 \)
\( \implies 9y = 20 + 7 \)
\( \implies 9y = 27 \)
\( \implies y = \frac{27}{9} \)
\( \implies y = 3 \)
In simple words: Shift -7 to the other side to make it +7. Add them to get 27, then divide by 9 to get 3.

Exam Tip: Practice quick mental addition and division to solve these simple equations rapidly.

Question 31. 2b - 14 = 8
Answer:
\( 2b - 14 = 8 \)
\( \implies 2b = 8 + 14 \)
\( \implies 2b = 22 \)
\( \implies b = \frac{22}{2} \)
\( \implies b = 11 \)
In simple words: Add 14 to 8 to get 22. Next, divide 22 by 2 to get the value of 11.

Exam Tip: Keep your steps neat to avoid sign calculation errors.

Question 32. \( \frac{7}{10}x + 6 = 41 \)
Answer:
\( \frac{7}{10}x + 6 = 41 \)
\( \implies \frac{7}{10}x = 41 - 6 \)
\( \implies \frac{7}{10}x = 35 \)
\( \implies 7x = 35 \times 10 \)
\( \implies 7x = 350 \)
\( \implies x = \frac{350}{7} \)
\( \implies x = 50 \)
In simple words: Subtract 6 from 41 to get 35. Multiply 35 by 10 to get 350, then divide by 7 to get 50.

Exam Tip: You can also divide 35 by 7 first, then multiply by 10 to simplify the calculation.

Question 33. \( \frac{5}{12}m - 12 = 48 \)
Answer:
\( \frac{5}{12}m - 12 = 48 \)
\( \implies \frac{5}{12}m = 48 + 12 \)
\( \implies \frac{5}{12}m = 60 \)
\( \implies 5m = 60 \times 12 \)
\( \implies 5m = 720 \)
\( \implies m = \frac{720}{5} \)
\( \implies m = 144 \)
In simple words: Add 12 to 48 to make 60. Then multiply 60 by 12 to get 720, and finally divide by 5 to get 144.

Exam Tip: Try to reduce the steps mentally if possible, or solve via \( m = 60 \times \frac{12}{5} \).

Exercise 12(B)

Question 1. 8y - 4y = 20
Answer:
\( 8y - 4y = 20 \)
\( \implies 4y = 20 \)
\( \implies y = \frac{20}{4} \)
\( \implies y = 5 \)
In simple words: Combine the y terms first. Eight y minus four y is four y. Then divide 20 by 4 to get 5.

Exam Tip: Collect and simplify like terms on the same side before doing any other operations.

Question 2. 9b - 4b + 3b = 16
Answer:
\( 9b - 4b + 3b = 16 \)
\( \implies 8b = 16 \)
\( \implies b = \frac{16}{8} \)
\( \implies b = 2 \)
In simple words: Simplify the b terms on the left: nine minus four is five, plus three is eight. Now divide 16 by 8 to get 2.

Exam Tip: Perform addition and subtraction of like terms from left to right carefully.

Question 3. 5y + 8 = 8y - 18
Answer:
\( 5y + 8 = 8y - 18 \)
\( \implies 5y - 8y = -18 - 8 \)
\( \implies -3y = -26 \)
\( \implies y = \frac{-26}{-3} \)
\( \implies y = \frac{26}{3} = 8\frac{2}{3} \)
In simple words: Group the y terms on the left and numbers on the right. This gives minus three y equals minus twenty-six. Dividing both sides by minus three gives twenty-six thirds, or eight and two-thirds.

Exam Tip: Group variables on one side and constants on the other, keeping track of negative signs.

Question 4. 6 = 7 + 2p - 5
Answer:
\( 6 = 7 + 2p - 5 \)
\( \implies -2p = 7 - 5 - 6 \)
\( \implies -2p = -4 \)
\( \implies p = \frac{-4}{-2} \)
\( \implies p = 2 \)
In simple words: Move 2p to the left to make it negative, and move 6 to the right. This gives minus two p equals minus four, so p is 2.

Exam Tip: You can also simplify the right side first: \( 6 = 2 + 2p \), then solve to save steps.

Question 5. 8 - 7x = 13x + 8
Answer:
\( 8 - 7x = 13x + 8 \)
\( \implies -7x - 13x = 8 - 8 \)
\( \implies -20x = 0 \)
\( \implies x = \frac{0}{-20} \)
\( \implies x = 0 \)
In simple words: Group the x terms to get minus twenty x, and the numbers to get zero. Dividing zero by any number gives zero.

Exam Tip: Zero divided by any non-zero number is always zero. Do not get confused by negative coefficients of zero.

Question 6. 4x - 5x + 2x = 28 + 3x
Answer:
\( 4x - 5x + 2x = 28 + 3x \)
\( \implies x = 28 + 3x \)
\( \implies x - 3x = 28 \)
\( \implies -2x = 28 \)
\( \implies x = \frac{28}{-2} \)
\( \implies x = -14 \)
In simple words: Combine the terms on the left to get x. Move 3x to the left to get minus two x equals 28, then divide 28 by minus two to get minus 14.

Exam Tip: Combine all terms on one side before doing transpositions across the equals sign.

Question 7. 9 + m = 6m + 8 - m
Answer:
\( 9 + m = 6m + 8 - m \)
\( \implies m - 6m + m = 8 - 9 \)
\( \implies 2m - 6m = -1 \)
\( \implies -4m = -1 \)
\( \implies m = \frac{-1}{-4} \)
\( \implies m = \frac{1}{4} \)
In simple words: Group all m terms on the left and numbers on the right. This gives minus four m equals minus one, which simplifies to one-quarter.

Exam Tip: Always simplify terms on each side first to reduce the chance of transposition errors.

Question 8. 24 = y + 2y + 3 + 4y
Answer:
\( 24 = y + 2y + 3 + 4y \)
\( \implies 24 - 3 = y + 2y + 4y \)
\( \implies 21 = 7y \)
\( \implies y = \frac{21}{7} \)
\( \implies y = 3 \)
In simple words: Subtract 3 from 24 to get 21. Combine the y terms to get seven y, then divide 21 by 7 to get 3.

Exam Tip: Combining like terms first makes the equation much easier to work with.

Question 9. 19x + 13 - 12x + 3 = 23
Answer:
\( 19x + 13 - 12x + 3 = 23 \)
\( \implies 19x - 12x = 23 - 13 - 3 \)
\( \implies 7x = 7 \)
\( \implies x = \frac{7}{7} \)
\( \implies x = 1 \)
In simple words: Group the x terms to get seven x. Group the numbers on the other side to get 7. Dividing 7 by 7 gives 1.

Exam Tip: Watch your signs carefully when moving multiple constant terms to the opposite side.

Question 10. 6b + 40 = -100 - b
Answer:
\( 6b + 40 = -100 - b \)
\( \implies 6b + b = -100 - 40 \)
\( \implies 7b = -140 \)
\( \implies b = \frac{-140}{7} \)
\( \implies b = -20 \)
In simple words: Move -b to the left to get seven b, and 40 to the right to get minus one hundred and forty. Dividing by 7 gives minus 20.

Exam Tip: When transposing positive 40 to a side with negative 100, they combine to become a larger negative number (-140).

Question 11. 6 - 5m - 1 + 3m = 0
Answer:
\( 6 - 5m - 1 + 3m = 0 \)
\( \implies -5m + 3m = -6 + 1 \)
\( \implies -2m = -5 \)
\( \implies m = \frac{-5}{-2} \)
\( \implies m = \frac{5}{2} = 2\frac{1}{2} \)
In simple words: Keep the m terms on the left to get minus two m, and move numbers to the right to get minus five. Dividing both sides by minus two gives two and a half.

Exam Tip: It is often simpler to simplify on the same side first: \( 5 - 2m = 0 \), then solve to avoid dealing with extra signs.

Question 12. 0.4x - 1.2 = 0.3x + 0.6
Answer:
\( 0.4x - 1.2 = 0.3x + 0.6 \)
\( \implies 0.4x - 0.3x = 0.6 + 1.2 \)
\( \implies 0.1x = 1.8 \)
\( \implies \frac{1}{10}x = \frac{18}{10} \)
\( \implies x = \frac{18}{10} \times \frac{10}{1} \)
\( \implies x = 18 \)
In simple words: Group decimal x terms on the left to get 0.1x, and decimal numbers on the right to get 1.8. Multiplying both sides by 10 to clear decimals gives 18.

Exam Tip: Converting decimals to fractions or multiplying the whole equation by 10 makes solving decimal linear equations much simpler.

Question 13. \( 6(x + 4) = 36 \)
Answer:
Expanding the brackets on the left side:
\( 6(x + 4) = 36 \)
\( \implies 6x + 24 = 36 \)
Subtracting 24 from both sides:
\( \implies 6x = 36 - 24 \)
\( \implies 6x = 12 \)
Dividing by 6 to isolate \( x \):
\( \implies x = \frac{12}{6} \)
\( \implies x = 2 \)
In simple words: Multiply 6 with both terms inside the bracket. Then move 24 to the right side as a subtraction, and finally divide by 6 to find \( x \).

Exam Tip: Be sure to multiply the term outside the parentheses with all the terms inside, not just the first one.

Question 14. \( 9(a + 5) + 2 = 11 \)
Answer:
First, distribute the multiplication across the terms inside the parentheses:
\( 9(a + 5) + 2 = 11 \)
\( \implies 9a + 45 + 2 = 11 \)
Rearrange the terms by grouping the constants on the right-hand side:
\( \implies 9a = 11 - 45 - 2 \)
\( \implies 9a = 11 - 47 \)
\( \implies 9a = -36 \)
Divide both sides by 9 to solve for \( a \):
\( \implies a = \frac{-36}{9} \)
\( \implies a = -4 \)
In simple words: Expand the bracket by multiplying 9 by \( a \) and 5. Group the numbers on the right side and divide by 9 to get \( -4 \).

Exam Tip: Keep careful track of negative signs when transferring terms to the other side of an equation.

Question 15. \( 4(x - 2) = 12 \)
Answer:
Begin by expanding the bracket on the left side:
\( 4(x - 2) = 12 \)
\( \implies 4x - 8 = 12 \)
Move the constant term to the right side of the equation:
\( \implies 4x = 12 + 8 \)
\( \implies 4x = 20 \)
Divide both sides of the equation by 4:
\( \implies x = \frac{20}{4} \)
\( \implies x = 5 \)
In simple words: Multiply 4 by both terms in the bracket. Move the \( -8 \) to the right side where it becomes \( +8 \), and then divide by 4.

Exam Tip: Alternatively, you can divide both sides by 4 first to get \( x - 2 = 3 \), which directly yields \( x = 5 \). This alternative approach can save time during an exam.

Question 16. \( -3(a - 6) = 24 \)
Answer:
Multiply the term outside by the terms in the bracket, noting that multiplying two negative values results in a positive:
\( -3(a - 6) = 24 \)
\( \implies -3a + 18 = 24 \)
Isolate the variable term by moving 18 to the right side:
\( \implies -3a = 24 - 18 \)
\( \implies -3a = 6 \)
Divide by -3 to find \( a \):
\( \implies a = \frac{6}{-3} \)
\( \implies a = -2 \)
In simple words: Multiply -3 by both terms inside the bracket, which gives \( -3a + 18 \). Move 18 to the right side and divide by -3.

Exam Tip: Pay close attention when a negative number is outside a bracket. A common mistake is writing \( -3a - 18 \) instead of \( -3a + 18 \).

Question 17. \( 7(x - 2) = 2(2x - 4) \)
Answer:
Expand the brackets on both sides of the equation:
\( 7(x - 2) = 2(2x - 4) \)
\( \implies 7x - 14 = 4x - 8 \)
Group the variable terms on the left and the constants on the right:
\( \implies 7x - 4x = -8 + 14 \)
\( \implies 3x = 6 \)
Divide by 3 to find \( x \):
\( \implies x = \frac{6}{3} \)
\( \implies x = 2 \)
In simple words: Multiply out the brackets on both sides of the equation. Group all \( x \) terms together on one side and the regular numbers on the other side.

Exam Tip: Try to move variables to the side where their coefficient remains positive to avoid dealing with negative divisions.

Question 18. \( (x - 4)(2x + 3) = 2x^2 \)
Answer:
Expand the expression on the left side by multiplying the two binomials:
\( (x - 4)(2x + 3) = 2x^2 \)
\( \implies x(2x + 3) - 4(2x + 3) = 2x^2 \)
\( \implies 2x^2 + 3x - 8x - 12 = 2x^2 \)
Subtract \( 2x^2 \) from both sides and combine the remaining terms:
\( \implies 2x^2 + 3x - 8x - 2x^2 = 12 \)
\( \implies -5x = 12 \)
Solve for \( x \) by dividing both sides by -5:
\( \implies x = \frac{12}{-5} \)
\( \implies x = -\frac{12}{5} \)
Convert the improper fraction into a mixed fraction:
\( \implies x = -2\frac{2}{5} \)
In simple words: Multiply the brackets on the left. The \( 2x^2 \) terms cancel out on both sides, which simplifies the equation into a basic linear one.

Exam Tip: When quadratic terms are present on both sides, they should cancel out. If they do not, go back and double-check your binomial expansion steps.

Question 19. \( 21 - 3(b - 7) = b + 20 \)
Answer:
Distribute -3 across the terms in the brackets:
\( 21 - 3(b - 7) = b + 20 \)
\( \implies 21 - 3b + 21 = b + 20 \)
Combine the constant terms on the left-hand side:
\( \implies -3b + 42 = b + 20 \)
Rearrange the terms to group variables on the left and constants on the right:
\( \implies -3b - b = 20 - 42 \)
\( \implies -4b = -22 \)
Divide both sides by -4 to solve for \( b \):
\( \implies b = \frac{-22}{-4} \)
\( \implies b = \frac{11}{2} \)
Convert the fraction to a mixed number:
\( \implies b = 5\frac{1}{2} \)
In simple words: Expand the bracket by multiplying -3 by both terms, making sure \( -3 \times -7 \) becomes \( +21 \). Then put all \( b \) terms on one side to solve.

Exam Tip: Be especially careful with signs when expanding a negative term outside a parenthesis, as this is a very common source of simple errors.

Question 20. \( x(x + 5) = x^2 + x + 32 \)
Answer:
Begin by expanding the expression on the left-hand side:
\( x(x + 5) = x^2 + x + 32 \)
\( \implies x^2 + 5x = x^2 + x + 32 \)
Bring all variable terms to the left side:
\( \implies x^2 + 5x - x^2 - x = 32 \)
The quadratic terms cancel out:
\( \implies 4x = 32 \)
Divide by 4 to solve for \( x \):
\( \implies x = \frac{32}{4} \)
\( \implies x = 8 \)
In simple words: Multiply \( x \) with the terms inside the bracket. The \( x^2 \) on both sides will cancel out, leaving a simple equation.

Exam Tip: Remember that once quadratic terms cancel out, the remaining linear equation can be solved by grouping like terms.

Exercise 12(C)

Question 1. \( \frac{x}{2} + x = 9 \)
Answer:
Express the second term with a denominator of 1:
\( \frac{x}{2} + \frac{x}{1} = 9 \)
Find a common denominator on the left side:
\( \frac{x + 2x}{2} = 9 \)
\( \implies \frac{3x}{2} = 9 \)
Multiply by 2 to clear the fraction:
\( \implies 3x = 9 \times 2 \)
\( \implies 3x = 18 \)
Divide by 3 to find \( x \):
\( \implies x = \frac{18}{3} \)
\( \implies x = 6 \)
In simple words: Add the half-x and full-x together to get \( \frac{3x}{2} \). Multiply by 2 and then divide by 3 to get 6.

Exam Tip: Expressing any standalone variable like \( x \) as \( \frac{x}{1} \) is a highly reliable way to avoid mistakes when adding fractions.

Question 2. \( \frac{x}{5} + 2x = 33 \)
Answer:
Write the second term as a fraction with a denominator of 1:
\( \frac{x}{5} + \frac{2x}{1} = 33 \)
Use a common denominator of 5 on the left side:
\( \frac{x + 10x}{5} = 33 \)
\( \implies \frac{11x}{5} = 33 \)
Multiply both sides by 5:
\( \implies 11x = 33 \times 5 \)
\( \implies 11x = 165 \)
Divide by 11 to solve for \( x \):
\( \implies x = \frac{165}{11} \)
\( \implies x = 15 \)
In simple words: Convert \( 2x \) into \( \frac{10x}{5} \) so that both terms have the same denominator. Add them together, multiply by 5, and divide by 11.

Exam Tip: You can simplify the step \( \frac{11x}{5} = 33 \) by dividing both sides by 11 first, which leaves \( \frac{x}{5} = 3 \) and simplifies the final calculation.

Question 3. \( \frac{3x}{4} + 4x = 38 \)
Answer:
Express the terms on the left side with a common denominator of 4:
\( \frac{3x}{4} + \frac{4x}{1} = 38 \)
\( \implies \frac{3x + 16x}{4} = 38 \)
\( \implies \frac{19x}{4} = 38 \)
Multiply both sides by 4 to remove the fraction:
\( \implies 19x = 38 \times 4 \)
\( \implies 19x = 152 \)
Divide by 19 to solve for \( x \):
\( \implies x = \frac{152}{19} \)
\( \implies x = 8 \)
In simple words: Convert \( 4x \) to \( \frac{16x}{4} \) to have a common denominator. Add them up, multiply by 4, and divide by 19.

Exam Tip: Knowing your mathematical tables helps. Recognizing that \( 19 \times 8 = 152 \) can speed up your division steps considerably.

Question 4. \( \frac{x}{2} + \frac{x}{5} = 14 \)
Answer:
Find the lowest common multiple (LCM) of 2 and 5, which is 10:
\( \frac{x}{2} + \frac{x}{5} = 14 \)
\( \implies \frac{5x + 2x}{10} = 14 \)
\( \implies \frac{7x}{10} = 14 \)
Multiply both sides by 10:
\( \implies 7x = 14 \times 10 \)
\( \implies 7x = 140 \)
Divide by 7 to find \( x \):
\( \implies x = \frac{140}{7} \)
\( \implies x = 20 \)
In simple words: Find a common denominator of 10. Combine the fractions, then multiply by 10 and divide by 7 to solve.

Exam Tip: Try simplifying equations before doing large multiplications. For instance, in \( \frac{7x}{10} = 14 \), dividing both sides by 7 gives \( \frac{x}{10} = 2 \) immediately.

Question 5. \( \frac{x}{3} - \frac{x}{4} = 2 \)
Answer:
Identify the common denominator for 3 and 4, which is 12:
\( \frac{x}{3} - \frac{x}{4} = 2 \)
\( \implies \frac{4x - 3x}{12} = 2 \)
Simplify the numerator on the left side:
\( \implies \frac{x}{12} = 2 \)
Multiply both sides by 12 to find \( x \):
\( \implies x = 2 \times 12 \)
\( \implies x = 24 \)
In simple words: Subtracting the fractions gives \( \frac{x}{12} = 2 \). Multiplying both sides by 12 yields the answer of 24.

Exam Tip: Always make sure you multiply both numerators correctly by their respective scaling factors when finding a common denominator.

Question 6. \( y + \frac{y}{2} = \frac{7}{4} - \frac{y}{4} \)
Answer:
Rewrite all terms with a common denominator of 4:
\( y + \frac{y}{2} = \frac{7}{4} - \frac{y}{4} \)
\( \implies \frac{4y}{4} + \frac{2y}{4} = \frac{7}{4} - \frac{y}{4} \)
Since all terms share the same denominator, we can equate the numerators:
\( \implies 4y + 2y = 7 - y \)
Group all terms with the variable \( y \) on the left side:
\( \implies 4y + 2y + y = 7 \)
\( \implies 7y = 7 \)
Divide by 7 to find \( y \):
\( \implies y = \frac{7}{7} \)
\( \implies y = 1 \)
In simple words: Give all terms the same denominator of 4, then remove the denominators completely and solve the remaining simple equation.

Exam Tip: When every single term on both sides has the same denominator, you can safely eliminate the denominators to simplify your working.

Question 7. \( \frac{4x}{3} - \frac{7x}{3} = 1 \)
Answer:
Since the denominators are already the same, subtract the numerators directly:
\( \frac{4x}{3} - \frac{7x}{3} = 1 \)
\( \implies \frac{4x - 7x}{3} = 1 \)
\( \implies \frac{-3x}{3} = 1 \)
Simplify the fraction:
\( \implies -x = 1 \)
Multiply by -1 to solve for \( x \):
\( \implies x = -1 \)
In simple words: Since the bottoms are both 3, just subtract the top numbers. This leaves \( -x = 1 \), which means \( x \) must be \( -1 \).

Exam Tip: Be careful with signs. Simplifying \( \frac{-3x}{3} \) results in \( -x \), and forgetting the negative sign is a very common error.

Question 8. \( \frac{1}{2}m + \frac{3}{4}m - m = 2.5 \)
Answer:
Write the terms on the left side using a common denominator of 4:
\( \frac{1}{2}m + \frac{3}{4}m - \frac{m}{1} = 2.5 \)
\( \implies \frac{2m + 3m - 4m}{4} = 2.5 \)
Multiply both sides by 4 to clear the fraction:
\( \implies 2m + 3m - 4m = 2.5 \times 4 \)
Simplify both sides of the equation:
\( \implies m = 10 \)
In simple words: Combine all terms on the left using a common denominator of 4. Multiply by 4 on the other side to find that \( m \) equals 10.

Exam Tip: Multiplying a decimal like 2.5 by an even integer like 4 often simplifies calculations by producing a clean whole number.

Question 9. \( \frac{2x}{3} + \frac{x}{2} - \frac{3x}{4} = 1 \)
Answer:
Find the LCM of the denominators 3, 2, and 4, which is 12:
\( \frac{2x}{3} + \frac{x}{2} - \frac{3x}{4} = 1 \)
\( \implies \frac{8x + 6x - 9x}{12} = 1 \)
Combine the terms in the numerator:
\( \implies \frac{5x}{12} = 1 \)
Multiply both sides by 12:
\( \implies 5x = 12 \)
Divide by 5 to find \( x \):
\( \implies x = \frac{12}{5} \)
Convert the improper fraction to a mixed number:
\( \implies x = 2\frac{2}{5} \)
In simple words: Use 12 as a common denominator to combine the fractions on the left side. Then, multiply by 12 and divide by 5 to solve.

Exam Tip: When the answer is an improper fraction, always check if your teacher or exam guidelines require converting it into a mixed fraction.

Question 10. \( \frac{3a}{4} + \frac{a}{6} = 66 \)
Answer:
Find a common denominator of 12 for the denominators 4 and 6:
\( \frac{3a}{4} + \frac{a}{6} = 66 \)
\( \implies \frac{9a + 2a}{12} = 66 \)
\( \implies \frac{11a}{12} = 66 \)
Multiply both sides by 12 to eliminate the denominator:
\( \implies 11a = 66 \times 12 \)
\( \implies 11a = 792 \)
Divide by 11 to solve for \( a \):
\( \implies a = \frac{792}{11} \)
\( \implies a = 72 \)
In simple words: Combine the fractions using 12 as a common denominator. Multiply both sides by 12, then divide by 11 to find the value of \( a \).

Exam Tip: Keep your numbers smaller by dividing before multiplying: \( \frac{11a}{12} = 66 \implies \frac{a}{12} = 6 \implies a = 72 \). This simplifies the math and helps prevent errors.

Question 11. \( \frac{2p}{3} - \frac{p}{5} = 35 \)
Answer:
Find a common denominator of 15 for 3 and 5:
\( \frac{2p}{3} - \frac{p}{5} = 35 \)
\( \implies \frac{10p - 3p}{15} = 35 \)
\( \implies \frac{7p}{15} = 35 \)
Multiply both sides by 15:
\( \implies 7p = 35 \times 15 \)
\( \implies 7p = 525 \)
Divide by 7 to solve for \( p \):
\( \implies p = \frac{525}{7} \)
\( \implies p = 75 \)
In simple words: Use 15 as the common denominator to combine the two fractions. Multiply both sides by 15, then divide by 7 to get 75.

Exam Tip: You can make the calculations easier by simplifying first: \( \frac{7p}{15} = 35 \implies \frac{p}{15} = 5 \implies p = 75 \). This is much quicker than multiplying large numbers.

Question 12. \( 0.6a + 0.2a = 0.4a + 8 \)
Answer:
Convert the decimal numbers into fractions:
\( 0.6a + 0.2a = 0.4a + 8 \)
\( \implies \frac{6}{10}a + \frac{2}{10}a = \frac{4}{10}a + \frac{8}{1} \)
Express both sides with a common denominator of 10:
\( \implies \frac{6a + 2a}{10} = \frac{4a + 80}{10} \ )
Multiply both sides by 10 to clear the denominators:
\( \implies 6a + 2 a = 4a + 80 \)
Combine the variable terms on the left-hand side:
\( \implies 8a = 4a + 80 \)
Subtract \( 4a \) from both sides:
\( \implies 8a - 4a = 80 \)
\( \implies 4a = 80 \)
Divide by 4 to solve for \( a \):
\( \implies a = \frac{80}{4} \)
\( \implies a = 20 \)
In simple words: Write decimals as fractions with a bottom of 10. Multiply the whole equation by 10 to remove the fractions, then solve for \( a \).

Exam Tip: Multiplying an entire equation with decimal values by 10 (or 100) is a highly effective way to eliminate decimals right at the start.

Question 13. \( p + 1.4p = 48 \)
Answer:
Convert the decimal term into a fraction:
\( p + 1.4p = 48 \)
\( \implies p + \frac{14}{10}p = 48 \)
Write both terms on the left side with a common denominator of 10:
\( \implies \frac{10p + 14p}{10} = 48 \)
\( \implies \frac{24p}{10} = 48 \)
Multiply both sides of the equation by 10:
\( \implies 24p = 480 \)
Divide by 24 to solve for \( p \):
\( \implies p = \frac{480}{24} \)
\( \implies p = 20 \)
In simple words: Change \( 1.4p \) to \( \frac{14}{10}p \). Combine them to get \( \frac{24p}{10} = 48 \), multiply by 10, and divide by 24.

Exam Tip: Alternatively, you can add them directly as decimals: \( 1p + 1.4p = 2.4p = 48 \implies p = \frac{48}{2.4} = 20 \). Choose the method that works best for you.

Question 14. \( 10\% \text{ of } x = 20 \)
Answer:
Convert the percentage into a fraction:
\( 10\% \text{ of } x = 20 \)
\( \implies \frac{10}{100} \times x = 20 \)
Simplify the fraction:
\( \implies \frac{x}{10} = 20 \)
Multiply by 10 to solve for \( x \):
\( \implies x = 20 \times 10 \)
\( \implies x = 200 \)
In simple words: Since 10% is the same as one-tenth, the equation means \( \frac{1}{10} \) of \( x \) is 20. Multiplying 20 by 10 gives 200.

Exam Tip: In math word problems, the word "of" indicates multiplication. Always convert percentages into fractions with a denominator of 100.

Question 15. \( y + 20\% \text{ of } y = 18 \)
Answer:
Convert the percentage into a fraction:
\( y + 20\% \text{ of } y = 18 \)
\( \implies y + \frac{20}{100} \times y = 18 \)
Express the left side with a common denominator of 100:
\( \implies \frac{100y + 20y}{100} = 18 \)
\( \implies \frac{120y}{100} = 18 \)
Multiply both sides by 100:
\( \implies 120y = 18 \times 100 \)
\( \implies 120y = 1800 \)
Divide by 120 to solve for \( y \):
\( \implies y = \frac{1800}{120} \)
\( \implies y = 15 \)
In simple words: Adding 20% of \( y \) to \( y \) is the same as 120% of \( y \). Convert this to a fraction and solve to find \( y = 15 \).

Exam Tip: Simplify the fraction first to keep calculations small. Reducing \( \frac{120}{100} \) to \( \frac{6}{5} \) turns the equation into \( \frac{6y}{5} = 18 \implies y = 15 \).

Question 16. \( x - 30\% \text{ of } x = 35 \)
Answer:
Express the percentage as a fraction:
\( x - 30\% \text{ of } x = 35 \)
\( \implies x - \frac{30}{100} \times x = 35 \)
Combine the terms using a common denominator of 100:
\( \implies \frac{100x - 30x}{100} = 35 \)
\( \implies \frac{70x}{100} = 35 \)
Multiply both sides by 100:
\( \implies 70x = 35 \times 100 \)
\( \implies 70x = 3500 \)
Divide by 70 to solve for \( x \):
\( \implies x = \frac{3500}{70} \)
\( \implies x = 50 \)
In simple words: Subtracting 30% from a whole leaves 70%. If 70% of a number is 35, then the full number is 50.

Exam Tip: Simplify your division by canceling out zeros from the numerator and denominator before completing the arithmetic.

Question 17. \( \frac{x+4}{2} + \frac{x}{3} = 7 \)
Answer:
Find a common denominator of 6 on the left side:
\( \frac{x+4}{2} + \frac{x}{3} = 7 \)
\( \implies \frac{3(x + 4) + 2x}{6} = 7 \)
Multiply both sides by 6 to remove the fraction:
\( \implies 3(x + 4) + 2x = 7 \times 6 \)
Expand the brackets and combine terms:
\( \implies 3x + 12 + 2x = 42 \)
\( \implies 5x + 12 = 42 \)
Subtract 12 from both sides:
\( \implies 5x = 42 - 12 \)
\( \implies 5x = 30 \)
Divide by 5 to find \( x \):
\( \implies x = \frac{30}{5} \)
\( \implies x = 6 \)
In simple words: Find a common denominator of 6, combine the terms on top, multiply both sides by 6, and then solve.

Exam Tip: Be sure to distribute the multiplication through the entire numerator. When multiplying \( x+4 \) by 3, both the \( x \) and the 4 must be multiplied.

Question 18. \( \frac{y+2}{3} + \frac{y+5}{4} = 6 \)
Answer:
Use a common denominator of 12 for the fractions on the left:
\( \frac{y+2}{3} + \frac{y+5}{4} = 6 \)
\( \implies \frac{4(y + 2) + 3(y + 5)}{12} = 6 \)
Multiply both sides by 12:
\( \implies 4(y + 2) + 3(y + 5) = 6 \times 12 \)
Expand the brackets on the left:
\( \implies 4y + 8 + 3y + 15 = 72 \)
Combine the variable and constant terms:
\( \implies 7y + 23 = 72 \)
Subtract 23 from both sides:
\( \implies 7y = 72 - 23 \)
\( \implies 7y = 49 \)
Divide by 7 to solve for \( y \):
\( \implies y = \frac{49}{7} \)
\( \implies y = 7 \)
In simple words: Find a common denominator of 12. Combine the numerators, multiply both sides by 12, then collect like terms to solve.

Exam Tip: It is always wise to double-check your final solution by substituting the value back into the original equation to verify that both sides are equal.

Question 19. \( \frac{3a-2}{7} - \frac{a-2}{4} = 2 \)
Answer:
Find a common denominator of 28 for the fractions on the left:
\( \frac{3a-2}{7} - \frac{a-2}{4} = 2 \)
\( \implies \frac{4(3a - 2) - 7(a - 2)}{28} = 2 \)
Multiply both sides by 28:
\( \implies 4(3a - 2) - 7(a - 2) = 2 \times 28 \)
Expand the terms, taking care with negative signs:
\( \implies 12a - 8 - 7a + 14 = 56 \)
Combine the variable and constant terms on the left:
\( \implies 5a + 6 = 56 \)
Subtract 6 from both sides:
\( \implies 5a = 56 - 6 \)
\( \implies 5a = 50 \)
Divide by 5 to find \( a \):
\( \implies a = \frac{50}{5} \)
\( \implies a = 10 \)
In simple words: Find a common denominator of 28. Be careful when expanding the second bracket: \( -7 \times -2 \) becomes \( +14 \). Then solve the simplified equation.

Exam Tip: Watch out for negative signs in front of a fraction. The negative sign applies to every term in that numerator when you expand it.

Question 20. \( \frac{1}{2}(x+5) - \frac{1}{3}(x-2) = 4 \)
Answer:
Write the equation with a common denominator of 6 on the left:
\( \frac{1}{2}(x+5) - \frac{1}{3}(x-2) = 4 \)
\( \implies \frac{3(x + 5) - 2(x - 2)}{6} = 4 \)
Multiply both sides by 6 to clear the fraction:
\( \implies 3(x + 5) - 2(x - 2) = 4 \times 6 \)
Expand the brackets:
\( \implies 3x + 15 - 2x + 4 = 24 \)
Combine terms on the left side:
\( \implies x + 19 = 24 \)
Subtract 19 from both sides to find \( x \):
\( \implies x = 24 - 19 \)
\( \implies x = 5 \)
In simple words: Set the fractions over a common denominator of 6. Clear it by multiplying the right side by 6, and then expand and solve.

Exam Tip: Pay special attention to sign changes during expansion. For example, \( -2(x - 2) \) expands to \( -2x + 4 \) instead of \( -2x - 4 \).

Question 21. \( \frac{x-1}{2} - \frac{x-2}{3} - \frac{x-3}{4} = 0 \)
Answer:
Find the lowest common multiple of 2, 3, and 4, which is 12:
\( \frac{x-1}{2} - \frac{x-2}{3} - \frac{x-3}{4} = 0 \)
\( \implies \frac{6(x - 1) - 4(x - 2) - 3(x - 3)}{12} = 0 \)
Multiply both sides by 12:
\( \implies 6(x - 1) - 4(x - 2) - 3(x - 3) = 0 \)
Expand the brackets carefully, keeping track of negative signs:
\( \implies 6x - 6 - 4x + 8 - 3x + 9 = 0 \)
Group the variable and constant terms:
\( \implies (6x - 4x - 3x) + (-6 + 8 + 9) = 0 \)
\( \implies -x + 11 = 0 \)
Isolate \( x \):
\( \implies -x = -11 \)
\( \implies x = 11 \)
In simple words: Combine fractions using a common denominator of 12. Expand brackets carefully, group terms, and solve for \( x \).

Exam Tip: Sign errors are very common when multiple negative signs are present. Work slowly when expanding terms like \( -4(x-2) \) and \( -3(x-3) \).

Question 22. \( \frac{x + 1}{3} + \frac{x + 4}{5} = \frac{x - 4}{7} \)
Answer:
Find the lowest common multiple (LCM) of the denominators on both sides, which are 3, 5, and 7. The LCM is 105.
Multiply all terms by 105 to clear the denominators:
\( 35(x + 1) + 21(x + 4) = 15(x - 4) \)
Now expand the brackets:
\( 35x + 35 + 21x + 84 = 15x - 60 \)
Combine the like terms on the left side:
\( 56x + 119 = 15x - 60 \)
Move the variable terms to the left side and constant numbers to the right side:
\( 56x - 15x = -60 - 119 \)
\( 41x = -179 \)
Divide by 41 to solve for x:
\( x = \frac{-179}{41} \)
Convert this improper fraction into a mixed number:
\( x = -4\frac{15}{41} \)
In simple words: Find a common denominator of 105 for all fractions. Multiply both sides by 105 to get rid of the division, open the brackets, and solve for x.

Exam Tip: Be very careful with negative signs when moving terms across the equals sign. Changing positive terms to negative ones incorrectly is a common source of mistakes.

Question 23. \( 15 - 2(5 - 3x) = 4(x - 3) + 13 \)
Answer:
Start by expanding the terms inside the brackets:
\( 15 - 10 + 6x = 4x - 12 + 13 \)
Simplify both sides:
\( 5 + 6x = 4x + 1 \)
Rearrange the equation by bringing variables to the left side and constant numbers to the right side:
\( 6x - 4x = 1 - 5 \)
\( 2x = -4 \)
Divide both sides by 2 to solve for x:
\( x = \frac{-4}{2} \)
\( x = -2 \)
Thus, the value of x is -2.
In simple words: Open the brackets first by multiplying the outer number with the inner terms. Then group the letters on one side and the numbers on the other side to find the answer.

Exam Tip: When multiplying -2 by -3x, the product is a positive 6x. Watch out for negative signs when expanding brackets, as this is where most mistakes happen.

Question 24. \( \frac{2x + 1}{3x - 2} = 1\frac{1}{4} \)
Answer:
First, convert the mixed fraction on the right side into an improper fraction:
\( 1\frac{1}{4} = \frac{5}{4} \)
This gives us the equation:
\( \frac{2x + 1}{3x - 2} = \frac{5}{4} \)
Use cross-multiplication to clear the fractions:
\( 4(2x + 1) = 5(3x - 2) \)
Multiply the terms inside the brackets:
\( 8x + 4 = 15x - 10 \)
Group the variable terms together and the numbers together:
\( 4 + 10 = 15x - 8x \)
\( 14 = 7x \)
Divide both sides by 7 to solve for x:
\( x = \frac{14}{7} \)
\( x = 2 \)
So, the answer is x = 2.
In simple words: Turn the mixed number into a fraction first. Then multiply diagonally to clear the division, and solve the simple equation that is left.

Exam Tip: When cross-multiplying, make sure to multiply the entire numerator by the opposite denominator. Do not leave out any terms inside the brackets.

Question 25. \( 21 - 3(x - 7) = x + 20 \)
Answer:
First, expand the expression by multiplying -3 with the terms inside the brackets:
\( 21 - 3x + 21 = x + 20 \)
Combine the constant numbers on the left side:
\( 42 - 3x = x + 20 \)
Shift the variable terms to the right side and constant numbers to the left side:
\( 42 - 20 = x + 3x \)
\( 22 = 4x \)
\( 4x = 22 \)
Divide both sides by 4 to find x:
\( x = \frac{22}{4} \)
Simplify the fraction by dividing both numbers by 2:
\( x = \frac{11}{2} \)
Change the improper fraction into a mixed number:
\( x = 5\frac{1}{2} \)
Therefore, the value of x is 5 1/2.
In simple words: Expand the brackets, simplify the numbers on each side, and move the x terms together. Write your final fraction as a mixed number.

Exam Tip: Remember that multiplying two negative numbers gives a positive result, so -3 times -7 becomes +21.

Question 26. \( \frac{3x - 2}{7} - \frac{x - 2}{4} = 2 \)
Answer:
To solve this, we find the LCM of the denominators, 7 and 4, which is 28.
Express the left side with a single common denominator:
\( \frac{4(3x - 2) - 7(x - 2)}{28} = 2 \)
Expand the terms in the numerator:
\( \frac{12x - 8 - 7x + 14}{28} = 2 \)
Combine like terms in the numerator:
\( \frac{5x + 6}{28} = 2 \)
Multiply both sides by 28 to remove the fraction:
\( 5x + 6 = 2 \times 28 \)
\( 5x + 6 = 56 \)
Subtract 6 from both sides of the equation:
\( 5x = 56 - 6 \)
\( 5x = 50 \)
Divide by 5 to solve for x:
\( x = \frac{50}{5} \)
\( x = 10 \)
Thus, x equals 10.
In simple words: Find a common denominator of 28 for both fractions. Subtract the numerators carefully, especially with the negative signs, and solve for x.

Exam Tip: When expanding -7(x - 2), be very careful to write it as -7x + 14. Forgetting to change the sign of the second term is a very common mistake.

Question 27. \( \frac{2x - 3}{3} - (x - 5) = \frac{x}{3} \)
Answer:
Write the equation down:
\( \frac{2x - 3}{3} - (x - 5) = \frac{x}{3} \)
To eliminate the fractions, multiply every term in the equation by 3:
\( 2x - 3 - 3(x - 5) = x \)
Expand the term inside the brackets:
\( 2x - 3 - 3x + 15 = x \)
Combine the like terms on the left side:
\( -x + 12 = x \ )
Add x to both sides to gather all variables on the right:
\( 12 = x + x \)
\( 2x = 12 \)
Divide both sides by 2:
\( x = \frac{12}{2} \)
\( x = 6 \)
So, the solution is x = 6.
In simple words: Multiply every part of the equation by 3 to clear the fractions. Then expand the brackets, simplify the terms, and find x.

Exam Tip: Do not forget to multiply the standalone bracket (x - 5) by 3 when clearing the denominators.

Question 28. \( \frac{x - 4}{7} = \frac{x + 3}{7} + \frac{x + 4}{5} \)
Answer:
Let the given equation be:
\( \frac{x - 4}{7} = \frac{x + 3}{7} + \frac{x + 4}{5} \)
Find the LCM of the denominators 7 and 5, which is 35.
Multiply all terms in the equation by 35 to clear the denominators:
\( 5(x - 4) = 5(x + 3) + 7(x + 4) \)
Open the brackets by multiplying the terms:
\( 5x - 20 = 5x + 15 + 7x + 28 \)
Group all terms containing x on the left side and constant numbers on the right side:
\( 5x - 5x - 7x = 15 + 28 + 20 \)
Combine the terms on both sides:
\( -7x = 63 \)
Divide by -7 to find the value of x:
\( x = \frac{63}{-7} \)
\( x = -9 \)
Hence, x is equal to -9.
In simple words: Multiply the entire equation by 35 to get rid of the denominators. Then expand the brackets, collect the x terms on one side, and solve.

Exam Tip: When moving terms, make sure to change their signs correctly. Double-check your final division step when dividing a positive number by a negative number.

Question 29. \( \frac{x - 1}{5} - \frac{x}{3} = 1 - \frac{x - 2}{2} \)
Answer:
The equation is:
\( \frac{x - 1}{5} - \frac{x}{3} = 1 - \frac{x - 2}{2} \)
The lowest common multiple of 5, 3, and 2 is 30.
Multiply each term by 30 to clear the fractions:
\( 6(x - 1) - 10x = 30 - 15(x - 2) \)
Expand the brackets on both sides:
\( 6x - 6 - 10x = 30 - 15x + 30 \)
Simplify the terms on both sides:
\( -4x - 6 = 60 - 15x \)
Move all terms with x to the left side and constant numbers to the right side:
\( 15x - 4x = 60 + 6 \)
\( 11x = 66 \)
Divide both sides by 11 to solve for x:
\( x = \frac{66}{11} \)
\( x = 6 \)
Therefore, the value of x is 6.
In simple words: Clear the denominators by multiplying everything by 30. Expand the expressions carefully, simplify both sides, and solve for x.

Exam Tip: Be careful with the term -15(x - 2). When you expand it, it becomes -15x + 30. A common error is writing -30 instead of +30.

Question 30. \( 2x + 20\% \text{ of } x = 12.1 \)
Answer:
Start with the equation:
\( 2x + 20\% \text{ of } x = 12.1 \)
Convert the percentage term into a fraction:
\( 2x + \frac{20}{100}x = 12.1 \)
Simplify the fraction:
\( 2x + \frac{2}{10}x = 12.1 \)
\( 2x + \frac{x}{5} = 12.1 \)
Take the LCM of the denominators to write the left side as a single fraction:
\( \frac{10x + x}{5} = 12.1 \ )
\( \frac{11x}{5} = 12.1 \)
Multiply both sides by 5:
\( 11x = 12.1 \times 5 \)
\( 11x = 60.5 \)
Divide by 11 to solve for x:
\( x = \frac{60.5}{11} \)
\( x = 5.5 \)
This can also be written as:
\( x = 5\frac{1}{2} \)
In simple words: Convert 20% to a simple fraction or decimal first. Add it to 2x, and then divide 12.1 by that total to find x.

Exam Tip: You can write percentages as decimals to make calculation faster. For instance, 20% of x is 0.2x, so the equation simplifies directly to 2.2x = 12.1.

Exercise 12(D)

Question 1. One-fifth of a number is 5, find the number.
Answer:
Let us assume the required number is x.
Based on the given condition:
\( \frac{1}{5}x = 5 \)
Multiply both sides by 5 to solve for x:
\( x = 5 \times 5 \)
\( x = 25 \)
Hence, the number is 25.
In simple words: We set up an equation where one-fifth of our unknown number is 5, then multiply 5 by 5 to find the total number.

Exam Tip: Always define your unknown quantity with a variable like x before writing down the equation.

Question 2. Six times a number is 72, find the number.
Answer:
Let the unknown number be x.
According to the problem statement:
\( 6x = 72 \)
Divide both sides by 6:
\( x = \frac{72}{6} \)
\( x = 12 \)
Therefore, the required number is 12.
In simple words: Six times a number means multiplying it by 6. To find the starting number, we divide 72 by 6.

Exam Tip: Remember that "times" means multiplication, and to undo it, we use division on both sides of the equation.

Question 3. If 15 is added to a number, the result is 69, find the number.
Answer:
Let the number we want to find be x.
From the given statement, we write:
\( x + 15 = 69 \)
Subtract 15 from both sides to find x:
\( x = 69 - 15 \)
\( x = 54 \)
Thus, the number is 54.
In simple words: If adding 15 to a number gives 69, we can work backwards by subtracting 15 from 69 to find the number.

Exam Tip: Subtract the given number from the total to easily find the unknown value.

Question 4. The sum of twice a number and 4 is 80, find the number.
Answer:
Let the number be represented by x.
According to the question:
\( 2x + 4 = 80 \)
Subtract 4 from both sides:
\( 2x = 80 - 4 \)
\( 2x = 76 \)
Divide both sides by 2:
\( x = \frac{76}{2} \)
\( x = 38 \)
So, the number is 38.
In simple words: Twice a number plus 4 equals 80. Subtract 4 from 80 first, then divide by 2 to get the number.

Exam Tip: Always perform the addition or subtraction steps before doing the division steps when solving linear equations.

Question 5. The difference between a number and one-fourth of itself is 24, find the number.
Answer:
Let us assume the number is x.
According to the given condition:
\( x - \frac{1}{4}x = 24 \)
Find a common denominator on the left side:
\( \frac{4x - x}{4} = 24 \)
\( \frac{3x}{4} = 24 \)
Multiply both sides by 4 and divide by 3:
\( x = 24 \times \frac{4}{3} \)
\( x = 8 \times 4 \)
\( x = 32 \)
Therefore, the required number is 32.
In simple words: Subtracting a quarter of a number from the whole number leaves three-quarters of it. If three-quarters is 24, the whole number must be 32.

Exam Tip: Make sure to express "one-fourth of itself" as \( \frac{1}{4}x \) (with the variable x included), not just as the constant fraction \( \frac{1}{4} \).

Question 6. Find a number whose one-third part exceeds its one-fifth part by 20.
Answer:
Let the unknown number be x.
As per the given condition:
\( \frac{1}{3}x - \frac{1}{5}x = 20 \)
The lowest common multiple of 3 and 5 is 15.
Write the fractions with the common denominator:
\( \frac{5x - 3x}{15} = 20 \)
\( \frac{2x}{15} = 20 \)
Multiply both sides by 15 and divide by 2:
\( x = \frac{20 \times 15}{2} \)
\( x = 10 \times 15 \)
\( x = 150 \)
Hence, the number is 150.
In simple words: One-third of a number is larger than one-fifth of it by 20. We subtract the two fractions to set up our equation and solve.

Exam Tip: When dealing with fractional parts of a number, always use the LCM of the denominators to simplify the equation.

Question 7. A number is as much greater than 35 as is less than 53. Find the number.
Answer:
Let the required number be x.
According to the problem:
\( x - 35 = 53 - x \)
Rearrange the equation by bringing both variables to one side and constants to the other:
\( x + x = 53 + 35 \)
\( 2x = 88 \)
Divide both sides by 2:
\( x = \frac{88}{2} \)
\( x = 44 \)
So, the number is 44.
In simple words: The number lies exactly halfway between 35 and 53. Adding them together and dividing by 2 gives the middle number.

Exam Tip: Questions with "as much greater than A as is less than B" can always be set up as \( x - A = B - x \).

Question 8. The sum of two numbers is 18. If one is twice the other, find the numbers.
Answer:
Let us assume the two numbers are x and y.
According to the first condition:
\( x + y = 18 \) ... (i)
Since one number is double the other:
\( x = 2y \) ... (ii)
Substitute the value of x from equation (ii) into equation (i):
\( 2y + y = 18 \)
\( 3y = 18 \)
Divide both sides by 3 to find y:
\( y = \frac{18}{3} \)
\( y = 6 \)
Now, find x by putting the value of y back into equation (ii):
\( x = 2 \times 6 \)
\( x = 12 \)
Therefore, the two required numbers are 12 and 6.
In simple words: We have two numbers that add up to 18. One is twice the size of the other, which means we can split 18 into three equal parts of 6. One number is 6 and the other is 12.

Exam Tip: In problems with two variables, express one in terms of the other first. This simplifies the system of equations into a single equation with one variable.

Question 9. A number is 15 more than the other. The sum of of the two numbers is 195. Find the numbers.
Answer:
Let the first number be x and the second number be y.
According to the first condition:
\( x = y + 15 \) ... (i)
According to the second condition:
\( x + y = 195 \) ... (ii)
Substitute the expression for x from equation (i) into equation (ii):
\( (y + 15) + y = 195 \)
\( 2y + 15 = 195 \)
\( 2y = 195 - 15 \)
\( 2y = 180 \)
\( y = \frac{180}{2} \)
\( y = 90 \)
Now, find the value of x by putting y = 90 into equation (i):
\( x = 90 + 15 \)
\( x = 105 \)
Therefore, the two numbers are 105 and 90.
In simple words: One number is 15 bigger than the other, and together they make 195. Subtract 15 from 195, halve the result to get the smaller number, and add 15 back to find the larger one.

Exam Tip: You can check your answer by adding both numbers (105 + 90 = 195) and checking if the difference is 15 (105 - 90 = 15).

Question 10. The sum of three consecutive even numbers is 54. Find the numbers.
Answer:
Let the three consecutive even numbers be x, x + 2, and x + 4.
According to the problem:
\( x + (x + 2) + (x + 4) = 54 \)
Simplify the left side:
\( 3x + 6 = 54 \)
Subtract 6 from both sides of the equation:
\( 3x = 54 - 6 \)
\( 3x = 48 \)
Divide by 3 to find x:
\( x = \frac{48}{3} \)
\( x = 16 \)
Now, compute the three numbers:
First even number: \( x = 16 \)
Second even number: \( x + 2 = 18 \)
Third even number: \( x + 4 = 20 \)
So, the three consecutive even numbers are 16, 18, and 20.
In simple words: Even numbers go up by 2. If we add three consecutive even numbers, we get 54. Solving this gives us 16, 18, and 20.

Exam Tip: Consecutive even numbers are always represented as x, x + 2, and x + 4 because even numbers always have a gap of 2 between them.

Question 11. The sum of three consecutive odd numbers is 63. Find the numbers.
Answer:
Let the first odd number be x, the second be x + 2, and the third be x + 4.
According to the given condition:
\( x + (x + 2) + (x + 4) = 63 \)
Combine the like terms on the left:
\( 3x + 6 = 63 \)
Subtract 6 from both sides of the equation:
\( 3x = 63 - 6 \)
\( 3x = 57 \)
Divide both sides by 3 to solve for x:
\( x = \frac{57}{3} \)
\( x = 19 \)
Now we can determine the three numbers:
First odd number: \( 19 \)
Second odd number: \( 19 + 2 = 21 \)
Third odd number: \( 19 + 4 = 23 \)
Hence, the three consecutive odd numbers are 19, 21, and 23.
In simple words: Consecutive odd numbers also increase by 2 each time. When we set up and solve the equation, we find the numbers are 19, 21, and 23.

Exam Tip: Remember that consecutive odd numbers are also represented by x, x + 2, x + 4 because the gap between any two consecutive odd numbers is 2, just like even numbers.

Question 12. A man has Rs. x from which he spends Rs. 6. If twice of the money left with him is Rs. 86, find x.
Answer:
Let the initial amount of money be Rs. x.
The money left after spending Rs. 6 is \( x - 6 \).
According to the given condition, twice this remaining amount is Rs. 86:
\( 2(x - 6) = 86 \)
Divide both sides by 2:
\( x - 6 = \frac{86}{2} \)
\( x - 6 = 43 \)
Add 6 to both sides of the equation to find x:
\( x = 43 + 6 \)
\( x = 49 \)
Therefore, the value of x is 49.
In simple words: After spending Rs. 6, the man has some money left. Double that amount is Rs. 86, which means he had Rs. 43 left. Adding back the Rs. 6 he spent, he originally had Rs. 49.

Exam Tip: Always write the expression for the remaining money, (x - 6), in brackets before multiplying by 2.

Question 13. A man is four times as old as his son. After 20 years, he will be twice as old as his son at that time. Find their present ages.
Answer:
Let the son's current age be x years.
This means the father's current age is 4x years.
After 20 years:
Son's age will be \( (x + 20) \) years.
Father's age will be \( (4x + 20) \) years.
According to the problem:
\( 4x + 20 = 2(x + 20) \)
Expand the bracket on the right side:
\( 4x + 20 = 2x + 40 \)
Rearrange the equation by bringing the variable terms to the left side:
\( 4x - 2x = 40 - 20 \)
\( 2x = 20 \)
Divide by 2 to solve for x:
\( x = 10 \)
Therefore:
The current age of the son is 10 years.
The current age of the father is \( 4 \times 10 = 40 \) years.
In simple words: The father is 4 times older than his son now. In 20 years, he will only be twice as old. This means the son is currently 10, and the father is 40.

Exam Tip: Remember to add 20 to both the father's and the son's age when setting up the equation for their future ages.

Question 14. If 5 is subtracted from three times a number, the result is 16. Find the number.
Answer:
Let the required number be represented by x.
According to the problem statement:
\( 3x - 5 = 16 \)
Add 5 to both sides of the equation:
\( 3x = 16 + 5 \)
\( 3x = 21 \)
Divide by 3 to solve for x:
\( x = \frac{21}{3} \)
\( x = 7 \)
Thus, the number is 7.
In simple words: Three times a number minus 5 equals 16. If we add 5 back, we get 21. Dividing by 3 tells us the starting number is 7.

Exam Tip: Translating words like "three times" into 3x and "subtracted from" into - 5 makes it easy to set up your equation.

Question 15. Find three consecutive natural numbers such that the sum of the first and the second is 15 more than the third.
Answer:
Let the three consecutive natural numbers be x, x + 1, and x + 2.
According to the condition:
\( x + (x + 1) = (x + 2) + 15 \)
Simplify both sides:
\( 2x + 1 = x + 17 \)
Subtract x and 1 from both sides:
\( 2x - x = 17 - 1 \)
\( x = 16 \)
Now, find the three numbers:
First number: \( 16 \)
Second number: \( 16 + 1 = 17 \)
Third number: \( 16 + 2 = 18 \)
Therefore, the three consecutive natural numbers are 16, 17, and 18.
In simple words: Consecutive natural numbers increase by 1. By setting up the equation, we find that the three numbers are 16, 17, and 18.

Exam Tip: Consecutive natural numbers are represented by x, x + 1, and x + 2 because they go up by exactly 1.

Question 16. The difference between two numbers is 7. Six times the smaller plus the larger is 77. Find the numbers.
Answer:
Let the smaller number be x and the larger number be y.
According to the first condition:
\( y - x = 7 \) ... (i)
According to the second condition:
\( 6x + y = 77 \) ... (ii)
From equation (i), we can express y in terms of x:
\( y = x + 7 \) ... (iii)
Substitute this expression for y into equation (ii):
\( 6x + (x + 7) = 77 \)
\( 7x + 7 = 77 \)
Subtract 7 from both sides:
\( 7x = 77 - 7 \)
\( 7x = 70 \)
Divide by 7 to solve for x:
\( x = \frac{70}{7} \)
\( x = 10 \)
Now, find y by substituting x = 10 into equation (iii):
\( y = 10 + 7 \)
\( y = 17 \)
Thus, the smaller number is 10 and the larger number is 17.
In simple words: The two numbers have a difference of 7. By substituting the larger number as x + 7, we can find that the numbers are 10 and 17.

Exam Tip: Substitution is a reliable method for solving linear equations with two variables. Express one variable in terms of the other and substitute it.

Question 17. The length of a rectangular plot exceeds its breadth by 5 metre. If the perimeter of the plot is 142 metres, find the length and the breadth of the plot.
Answer:
Let the length of the rectangular plot be x meters and the breadth be y meters.
From the first condition:
\( x = y + 5 \) ... (i)
Since the perimeter of the rectangle is 142 meters:
\( 2(x + y) = 142 \)
Divide both sides by 2:
\( x + y = 71 \) ... (ii)
Substitute the value of x from equation (i) into equation (ii):
\( (y + 5) + y = 71 \)
\( 2y + 5 = 71 \)
Subtract 5 from both sides:
\( 2y = 71 - 5 \)
\( 2y = 66 \)
Divide by 2 to find y:
\( y = \frac{66}{2} \)
\( y = 33 \)
Now, find x by putting the value of y back into equation (i):
\( x = 33 + 5 \)
\( x = 38 \)
Hence, the length of the plot is 38 meters and the breadth is 33 meters.
In simple words: The length is 5 meters more than the breadth. Since the sum of length and breadth is half of the perimeter (71 meters), we solve to find they are 38 meters and 33 meters.

Exam Tip: Always write the units (like 'm' or 'meters') in your final answer when solving word problems involving measurements.

Question 18. The numerator of a fraction is four less than its denominator. If 1 is added to both, is numerator and denominator, the fraction becomes 1/2 Find the fraction.
Answer:
Let the numerator of the fraction be x and the denominator be y.
According to the first condition:
\( x = y - 4 \) ... (i)
According to the second condition:
\( \frac{x + 1}{y + 1} = \frac{1}{2} \)
Cross-multiply to simplify:
\( 2(x + 1) = y + 1 \)
\( 2x + 2 = y + 1 \)
\( 2x - y = 1 - 2 \)
\( 2x - y = -1 \) ... (ii)
Substitute the value of x from equation (i) into equation (ii):
\( 2(y - 4) - y = -1 \)
\( 2y - 8 - y = -1 \)
\( y - 8 = -1 \)
\( y = -1 + 8 \)
\( y = 7 \)
Put this value of y back into equation (i) to find x:
\( x = 7 - 4 \)
\( x = 3 \)
So, the numerator is 3 and the denominator is 7.
Therefore, the required fraction is \( \frac{3}{7} \).
In simple words: The top number is 4 less than the bottom number. If we add 1 to both, the fraction becomes 1/2. Solving this tells us the fraction is 3/7.

Exam Tip: Always state the final fraction clearly as \( \frac{numerator}{denominator} \) at the end of your answer, rather than just leaving the values of x and y.

Question 19. A man is thrice as old as his son. After 12 years, he will be twice as old as his son at that time. Find their present ages.
Answer:
Let the present age of the son be x years.
This means the present age of the father is 3x years.
After 12 years:
Son's age will be \( (x + 12) \) years.
Father's age will be \( (3x + 12) \) years.
Based on the given condition:
\( 3x + 12 = 2(x + 12) \)
Expand the term on the right side:
\( 3x + 12 = 2x + 24 \)
Move all the terms with variables to the left and constants to the right:
\( 3x - 2x = 24 - 12 \)
\( x = 12 \)
Therefore:
The son's present age is 12 years.
The father's present age is \( 3 \times 12 = 36 \) years.
In simple words: The father is three times as old as his son today. In 12 years, he will be twice as old. This means the son is 12 and the father is 36.

Exam Tip: Be sure to multiply both terms inside the bracket by 2 when expanding 2(x + 12).

Question 20. A sum of Rs. 500 is in the form of notes of denominations of Rs. 5 and Rs. 10. If the total number of notes is 90, find the number of notes of each type.
Answer:
Let the number of Rs. 5 notes be x.
Since the total number of notes is 90, the number of Rs. 10 notes will be \( 90 - x \).
The value of the Rs. 5 notes is \( 5 \times x = 5x \).
The value of the Rs. 10 notes is \( 10(90 - x) = 900 - 10x \).
According to the problem, the total value of all notes is Rs. 500:
\( 5x + (900 - 10x) = 500 \ )
Simplify the expression:
\( 5x + 900 - 10x = 500 \)
\( -5x + 900 = 500 \)
Subtract 900 from both sides:
\( -5x = 500 - 900 \)
\( -5x = -400 \)
Divide by -5:
\( x = \frac{-400}{-5} \)
\( x = 80 \)
Therefore:
The number of Rs. 5 notes is 80.
The number of Rs. 10 notes is \( 90 - 80 = 10 \).
In simple words: If we have 90 notes in total, let's say we have x notes of Rs. 5. The rest are Rs. 10 notes. Solving the total value equation shows we have 80 notes of Rs. 5 and 10 notes of Rs. 10.

Exam Tip: Always verify your solution by calculating the total value: (80 × Rs. 5) + (10 × Rs. 10) = Rs. 400 + Rs. 100 = Rs. 500.