Selina Concise Solutions for ICSE Class 8 Mathematics Chapter 11 Algebraic Expressions

Exercise 11(A)

Question 1. Separate the constants and variables from the following : \( -7, 7 + x, 7x + yz, \sqrt{5}, \sqrt{xy}, \frac{3yz}{8}, 4.5y - 3x, 8 - 5, 8 - 5x, 8x - 5y \times p \text{ and } 3y^2z + 4x \)
Answer: Constants are terms whose values do not change: \( -7, \sqrt{5}, 8 - 5 \).
Variables are expressions that contain letters, as their values can change: \( 7 + x, 7x + yz, \sqrt{xy}, \frac{3yz}{8}, 4.5y - 3x, 8 - 5x, 8x - 5y \times p, 3y^2z + 4x \).
In simple words: Constants are numbers that stay the same. Variables are terms that have letters because their values can change.

Exam Tip: Remember that a constant cannot contain any variable letters. Expression elements like \( 8 - 5 \) simplify to a single number, so they are constants.

Question 2. Write the number of terms in each of the following polynomials.
(i) 5x2 + 3 x ax
(ii) ax ÷ 4 – 7
(iii) ax – by + y x z
(iv) 23 + a x b ÷ 2.
Answer: We can find the terms by simplifying the expressions first:
(i) \( 5x^2 + 3 \times ax = 5x^2 + 3ax \) - This expression has 2 terms.
(ii) \( ax \div 4 - 7 = \frac{ax}{4} - 7 \) - This expression has 2 terms.
(iii) \( ax - by + y \times z = ax - by + yz \) - This expression has 3 terms.
(iv) \( 23 + a \times b \div 2 = 23 + \frac{ab}{2} \) - This expression has 2 terms.
In simple words: To find the number of terms, first simplify any multiplication or division. Then, count each part separated by a plus or minus sign.

Exam Tip: Multiplication and division operations do not split terms - only addition and subtraction symbols do.

Question 3. Separate monomials, binomials, trinomials and polynomials from the following algebraic expressions : \( 8 - 3x, xy^2, 3y^2 - 5y + 8, 9x - 3x^2 + 15x^3 - 7, 3x \times 5y, 3x \div 5y, 2y \div 7 + 3x - 7 \text{ and } 4 - ax^2 + bx + y \)
Answer: Let us group the given expressions by how many terms they have after simplification:
Monomials (one term): \( xy^2, 3x \times 5y, 3x \div 5y \)
Binomials (two terms): \( 8 - 3x \)
Trinomials (three terms): \( 3y^2 - 5y + 8, 2y \div 7 + 3x - 7 \)
Polynomials (expressions with one or more terms): \( 8 - 3x, 3y^2 - 5y + 8, 9x - 3x^2 + 15x^3 - 7, 2y \div 7 + 3x - 7, 4 - ax^2 + bx + y \)
In simple words: Monomials have one term, binomials have two, and trinomials have three. Polynomials include any expression with one or more terms.

Exam Tip: Simplify multiplication and division inside terms before counting them to avoid mistakenly classifying them.

Question 4. Write the degree of each polynomial given below :
(i) xy + 7z
(ii) x2 – 6x3 + 8 y
(iii) y – 6y2 + 5y8
(iv) xyz – 3
(v) xy + yz2 – xz3
(vi) x5y7 – 8x3y8 + 10x4y4z4
Answer: The highest power of the variables in any single term determines the degree of that polynomial:
(i) For \( xy + 7z \), the degree is \( 1+1 = 2 \).
(ii) For \( x^2 - 6x^3 + 8y \), the highest power is \( 3 \), so the degree is \( 3 \).
(iii) For \( y - 6y^2 + 5y^8 \), the highest power is \( 8 \), so the degree is \( 8 \).
(iv) For \( xyz - 3 \), the term \( xyz \) has degree \( 1+1+1 = 3 \).
(v) For \( xy + yz^2 - xz^3 \), the highest sum of powers is in \( -xz^3 \) which is \( 1+3 = 4 \), so the degree is \( 4 \).
(vi) For \( x^5 y^7 - 8x^3 y^8 + 10x^4 y^4 z^4 \), the terms have degrees: \( 5+7=12 \), \( 3+8=11 \), and \( 4+4+4=12 \). Thus, the degree is \( 12 \).
In simple words: The degree is the biggest exponent on a letter in the expression. If a term has more than one letter, add their exponents together first.

Exam Tip: For terms with multiple variables like \( xyz \), always add the exponents of all the variables in that term to find its degree.

Question 5. Write the coefficient of :
(i) ab in 7abx ,
(ii) 7a in 7abx ;
(iii) 5x2 in 5x2 – 5x ;
(iv) 8 in a2 – 8ax + a ;
(v) 4xy in x2 – 4xy + y2.
Answer: Let us find the coefficient by dividing the term by the specified factor:
(i) In \( 7abx \), the coefficient of \( ab \) is \( 7x \).
(ii) In \( 7abx \), the coefficient of \( 7a \) is \( bx \).
(iii) In the term \( 5x^2 \), the coefficient of \( 5x^2 \) is \( 1 \).
(iv) In the term \( -8ax \), the coefficient of \( 8 \) is \( -ax \).
(v) In the term \( -4xy \), the coefficient of \( 4xy \) is \( -1 \).
In simple words: To find the coefficient of a specific part in a term, just remove that part. Whatever is left over is the coefficient.

Exam Tip: Keep the negative sign with the coefficient if the original term in the polynomial is negative.

Question 6. In \( \frac{5}{7} xy^2 z^3 \), write the coefficient of
(i) 5
(ii) \(\frac{5}{7}\)
(iii) 5x
(iv) xy2
(v) z3
(vi) xz3
(vii) 5xy2
(viii) \(\frac{1}{7}\) yz
(ix) z
(x) yz2
(xi) 5xyz
Answer: By removing the given term from \( \frac{5}{7} xy^2 z^3 \), we obtain its coefficient:
(i) For \( 5 \), the coefficient is \( \frac{1}{7} xy^2 z^3 \).
(ii) For \( \frac{5}{7} \), the coefficient is \( xy^2 z^3 \).
(iii) For \( 5x \), the coefficient is \( \frac{1}{7} y^2 z^3 \).
(iv) For \( xy^2 \), the coefficient is \( \frac{5}{7} z^3 \).
(v) For \( z^3 \), the coefficient is \( \frac{5}{7} xy^2 \).
(vi) For \( xz^3 \), the coefficient is \( \frac{5}{7} y^2 \).
(vii) For \( 5xy^2 \), the coefficient is \( \frac{1}{7} z^3 \).
(viii) For \( \frac{1}{7} yz \), the coefficient is \( 5xyz^2 \).
(ix) For \( z \), the coefficient is \( \frac{5}{7} xy^2 z^2 \).
(x) For \( yz^2 \), the coefficient is \( \frac{5}{7} xyz \).
(xi) For \( 5xyz \), the coefficient is \( \frac{1}{7} yz^2 \).
In simple words: The coefficient is what you get when you divide the whole term by the part you are asked about.

Exam Tip: Always ensure your final coefficient product multiplied by the given term equals the original expression exactly.

Question 7. In each polynomial, given below, separate the like terms :
(i) 3xy, – 4yx2, 2xy2, 2.5x2y, –8yx, –3.2y2x and x2y
(ii) y2z3, xy2z3, – 5x2yz, –4y2z3, –8xz3y2, 3x2yz and 2z3y2
Answer: Like terms are those terms that have the exact same variables raised to the exact same powers:
(i) The sets of like terms are:
- \( 3xy \text{ and } -8yx \)
- \( -4yx^2, 2.5x^2y, \text{ and } x^2y \)
- \( 2xy^2 \text{ and } -3.2y^2x \)
(ii) The sets of like terms are:
- \( y^2z^3, -4y^2z^3, \text{ and } 2z^3y^2 \)
- \( xy^2z^3 \text{ and } -8xz^3y^2 \)
- \( -5x^2yz \text{ and } 3x^2yz \)
In simple words: Like terms are terms that have the exact same letters and exponents. It does not matter what order the letters are written in.

Exam Tip: The order of variables does not affect like terms. For example, \( -4yx^2 \) and \( 2.5x^2y \) are like terms because the powers of \( x \) and \( y \) are identical.

Exercise 11(B)

Question 1. Evaluate :
(i) \(-7x^2 + 18x^2 + 3x^2 - 5x^2\)
(ii) \(b^2y - 9b^2y + 2b^2y - 5b^2y\)
(iii) \(abx - 15abx - 10abx + 32abx\)
(iv) \(7x - 9y + 3 - 3x - 5y + 8\)
(v) \(3x^2 + 5xy - 4y^2 + x^2 - 8xy - 5y^2\)
Answer: We can evaluate each expression by combining the like terms:
(i) Group the positive and negative terms:
\( -7x^2 + 18x^2 + 3x^2 - 5x^2 \)
\( = (18 + 3)x^2 - (7 + 5)x^2 \)
\( = 21x^2 - 12x^2 \)
\( = 9x^2 \)

(ii) Combine positive and negative terms:
\( b^2y - 9b^2y + 2b^2y - 5b^2y \)
\( = (1 + 2)b^2y - (9 + 5)b^2y \)
\( = 3b^2y - 14b^2y \)
\( = -11b^2y \)

(iii) Combine positive and negative coefficients:
\( abx - 15abx - 10abx + 32abx \)
\( = (1 + 32)abx - (15 + 10)abx \)
\( = 33abx - 25abx \)
\( = 8abx \)

(iv) Rearrange to bring like terms together:
\( 7x - 9y + 3 - 3x - 5y + 8 \)
\( = (7x - 3x) + (-9y - 5y) + (3 + 8) \)
\( = 4x - 14y + 11 \)

(v) Group identical terms:
\( 3x^2 + 5xy - 4y^2 + x^2 - 8xy - 5y^2 \)
\( = (3x^2 + x^2) + (5xy - 8xy) - (4y^2 + 5y^2) \)
\( = 4x^2 - 3xy - 9y^2 \)
In simple words: To solve these, put terms with the same letters next to each other. Then, add or subtract their numbers.

Exam Tip: Group positive terms together and negative terms together first to avoid simple calculation errors.

Question 2. Add :
(i) \( 5a + 3b, a - 2b, 3a + 5b \)
(ii) \( 8x - 3y + 7z, -4x + 5y - 4z, -x - y - 2z \)
(iii) \( 3b - 7c + 10, 5c - 2b - 15, 15 + 12c + b \)
(iv) \( a - 3b + 3 ; 2a + 5 - 3c ; 6c - 15 + 6b \)
(v) \( 13ab - 9cd - xy ; 5xy ; 15cd - 7ab ; 6xy - 3cd \)
(vi) \( x^3 - x^2y + 5xy^2 + y^3 ; -x^3 - 9xy^2 + y^3 ; 3x^2y + 9xy^2 \)
(vii) \( a^6 - 4a^4 + 6a ; 5a^6 + 5a^4 + 6a ; 12a^6 - 10a \)
(viii) \( 2ax - 6by + 4cz, 4by - 14ax, 9cz - 4ax - 6by \)
Answer: Let us add the expressions by arranging the like terms vertically in columns:
(i) Adding the terms:
\[ \begin{array}{rccc} & 5a & + & 3b \\ & a & - & 2b \\ + & 3a & + & 5b \\ \hline & 9a & + & 6b \end{array} \]

(ii) Adding the terms:
\[ \begin{array}{rccccc} & 8x & - & 3y & + & 7z \\ & -4x & + & 5y & - & 4z \\ + & -x & - & y & - & 2z \\ \hline & 3x & + & y & + & z \end{array} \]

(iii) Aligning the terms for \( b \), \( c \), and constants:
\[ \begin{array}{rccccc} & 3b & - & 7c & + & 10 \\ & -2b & + & 5c & - & 15 \\ + & b & + & 12c & + & 15 \\ \hline & 2b & + & 10c & + & 10 \end{array} \]

(iv) Aligning the terms for \( a \), \( b \), \( c \), and constants:
\[ \begin{array}{rcccccc} & a & - & 3b & & & + & 3 \\ & 2a & & & - & 3c & + & 5 \\ + & & + & 6b & + & 6c & - & 15 \\ \hline & 3a & + & 3b & + & 3c & - & 7 \end{array} \]

(v) Aligning the terms for \( ab \), \( cd \), and \( xy \):
\[ \begin{array}{rccccc} & 13ab & - & 9cd & - & xy \\ & & & & + & 5xy \\ & -7ab & + & 15cd & & \\ + & & - & 3cd & + & 6xy \\ \hline & 6ab & + & 3cd & + & 10xy \end{array} \]

(vi) Aligning the terms by power:
\[ \begin{array}{rccccccc} & x^3 & - & x^2y & + & 5xy^2 & + & y^3 \\ & -x^3 & & & - & 9xy^2 & + & y^3 \\ + & & + & 3x^2y & + & 9xy^2 & & \\ \hline & & & 2x^2y & + & 5xy^2 & + & 2y^3 \end{array} \]

(vii) Combining the coefficients of \( a^6 \), \( a^4 \), and \( a \):
\[ \begin{array}{rcccc} & a^6 & - & 4a^4 & + & 6a \\ & 5a^6 & + & 5a^4 & + & 6a \\ + & 12a^6 & & & - & 10a \\ \hline & 18a^6 & + & a^4 & + & 2a \end{array} \]

(viii) Grouping by \( ax \), \( by \), and \( cz \):
\[ \begin{array}{rccccc} & 2ax & - & 6by & + & 4cz \\ & -14ax & + & 4by & & \\ + & -4ax & - & 6by & + & 9cz \\ \hline & -16ax & - & 8by & + & 13cz \end{array} \]
In simple words: When adding expressions, put the terms with the same letters in vertical columns, and then add their numbers.

Exam Tip: Leave empty spaces or write zero when a column does not have a matching term for a particular variable.

Question 3. Find the total savings of a boy who saves Rs. (4x – 6y) ; Rs. (6x + 2y) ; Rs. (4y – x) and Rs. (y – 2x) for four consecutive weeks.
Answer: To find the total amount saved, we sum the savings from all four weeks:
\[ \begin{array}{rcc} & 4x & - & 6y \\ & 6x & + & 2y \\ & -x & + & 4y \\ + & -2x & + & y \\ \hline & 7x & + & y \end{array} \]
Total savings = Rs. \( (7x + y) \)
In simple words: To find the total savings, write all four weekly amounts in columns and add the numbers with the same letters together.

Exam Tip: Do not forget to write the final units like Rs. clearly in your answer statement.

Question 4. Subtract :
(i) 4xy2 from 3xy2 ;
(ii) -2x2y + 3xy2 from 8x2y ;
(iii) 3a - 5b + c + 2d from 7a - 3b + c - 2d
(iv) x3 - 4x - 1 from 3x3 - x2 + 6
(v) 6a + 3 from a3 - 3a2 + 4a + 1
(vi) cab - 4cad - cbd from 3abc + 5bcd - cda
(vii) a2 + ab + b2 from 4a2 - 3ab + 2b2.
Answer: To subtract, we write the expression to be subtracted below the other expression and invert all signs of the bottom terms before adding:
(i) Direct subtraction:
\( 3xy^2 - 4xy^2 = -xy^2 \)

(ii) Writing in columns:
\[ \begin{array}{rcc} & 8x^2y & \\ & -2x^2y & + & 3xy^2 \\ \text{Change signs:} & (+) & & (-) \\ \hline & 10x^2y & - & 3xy^2 \end{array} \]

(iii) Aligning and changing signs:
\[ \begin{array}{rccccccc} & 7a & - & 3b & + & c & - & 2d \\ & 3a & - & 5b & + & c & + & 2d \\ \text{Change signs:} & (-) & & (+) & & (-) & & (-) \\ \hline & 4a & + & 2b & & & - & 4d \end{array} \]

(iv) Aligning matching powers and changing signs:
\[ \begin{array}{rcccccc} & 3x^3 & - & x^2 & & & + & 6 \\ & x^3 & & & - & 4x & - & 1 \\ \text{Change signs:} & (-) & & & & (+) & & (+) \\ \hline & 2x^3 & - & x^2 & + & 4x & + & 7 \end{array} \]

(v) Placing like terms under each other:
\[ \begin{array}{rccccccc} & a^3 & - & 3a^2 & + & 4a & + & 1 \\ & & & & + & 6a & + & 3 \\ \text{Change signs:} & & & & & (-) & & (-) \\ \hline & a^3 & - & 3a^2 & - & 2a & - & 2 \end{array} \]

(vi) Rewriting terms like \( cab \) as \( abc \), \( cbd \) as \( bcd \), and \( cad \) as \( cda \) before subtracting:
\[ \begin{array}{rccccc} & 3abc & + & 5bcd & - & cda \\ & abc & - & bcd & - & 4cda \\ \text{Change signs:} & (-) & & (+) & & (+) \\ \hline & 2abc & + & 6bcd & + & 3cda \end{array} \]

(vii) Column subtraction:
\[ \begin{array}{rccccc} & 4a^2 & - & 3ab & + & 2b^2 \\ & a^2 & + & ab & + & b^2 \\ \text{Change signs:} & (-) & & (-) & & (-) \\ \hline & 3a^2 & - & 4ab & + & b^2 \end{array} \]
In simple words: When subtracting, put the second expression below the first, change all the signs of the bottom terms, and then add them up.

Exam Tip: The word "from" means the second expression goes on top. Be extremely careful to invert every single sign of the expression being subtracted.

Question 5. (i) Take away -3x3 + 4x2 - 5x+ 6 from 3x3 - 4x2 + 5x - 6
(ii) Take m2 + m + 4 from -m2 + 3m + 6 and the result from m2 + m + 1.
Answer: We will solve both parts using column subtraction:
(i) Subtracting the first expression from the second:
\[ \begin{array}{rccccccc} & 3x^3 & - & 4x^2 & + & 5x & - & 6 \\ & -3x^3 & + & 4x^2 & - & 5x & + & 6 \\ \text{Change signs:} & (+) & & (-) & & (+) & & (-) \\ \hline & 6x^3 & - & 8x^2 & + & 10x & - & 12 \end{array} \]

(ii) First, subtract \( m^2 + m + 4 \) from \( -m^2 + 3m + 6 \):
\[ \begin{array}{rccccc} & -m^2 & + & 3m & + & 6 \\ & m^2 & + & m & + & 4 \\ \text{Change signs:} & (-) & & (-) & & (-) \\ \hline & -2m^2 & + & 2m & + & 2 \end{array} \]
Now, subtract this result from \( m^2 + m + 1 \):
\[ \begin{array}{rccccc} & m^2 & + & m & + & 1 \\ & -2m^2 & + & 2m & + & 2 \\ \text{Change signs:} & (+) & & (-) & & (-) \\ \hline & 3m^2 & - & m & - & 1 \end{array} \]
In simple words: For the second part, subtract the first expression from the second to get a result, and then subtract that result from the final expression.

Exam Tip: Break down multi-step word problems into individual operations to prevent sign errors.

Question 6. Subtract the sum of 5y2 + y - 3 and y2 - 3y + 7 from 6y2 + y - 2.
Answer: First, find the sum of the two given expressions:
\[ \begin{array}{rcccc} & 5y^2 & + & y & - & 3 \\ + & y^2 & - & 3y & + & 7 \\ \hline & 6y^2 & - & 2y & + & 4 \end{array} \]
Next, subtract this sum from \( 6y^2 + y - 2 \):
\[ \begin{array}{rcccc} & 6y^2 & + & y & - & 2 \\ & 6y^2 & - & 2y & + & 4 \\ \text{Change signs:} & (-) & & (+) & & (-) \\ \hline & & & 3y & - & 6 \end{array} \]
In simple words: First, add the two expressions together. Then, subtract this sum from the final expression.

Exam Tip: Use brackets around the sum when subtracting it from the final expression to ensure the negative sign is distributed correctly.

Question 7. What must be added to x4 - x3 + x2 + x + 3 to obtain x4 + x2 - 1 ?
Answer: To find the required expression, we subtract the first expression from the second:
\[ \begin{array}{rccccccccc} & x^4 & & & + & x^2 & & & - & 1 \\ & x^4 & - & x^3 & + & x^2 & + & x & + & 3 \\ \text{Change signs:} & (-) & & (+) & & (-) & & (-) & & (-) \\ \hline & & & x^3 & & & - & x & - & 4 \end{array} \]
Thus, the expression that needs to be added is \( x^3 - x - 4 \).
In simple words: To find what must be added, subtract the first expression from the second one.

Exam Tip: Align the terms carefully under their matching powers. If a power is missing in one expression, leave a blank space.

Question 8. (i) How much more than 2x2 + 4xy + 2y2 is 5x2 + 10xy - y2 ?
(ii) How much less 2a2 + 1 is than 3a2 - 6 ?
Answer: We find these differences by performing subtraction:
(i) To find how much more the second expression is, we subtract the first expression from it:
\[ \begin{array}{rccccc} & 5x^2 & + & 10xy & - & y^2 \\ & 2x^2 & + & 4xy & + & 2y^2 \\ \text{Change signs:} & (-) & & (-) & & (-) \\ \hline & 3x^2 & + & 6xy & - & 3y^2 \end{array} \]

(ii) To find how much less the first expression is, we subtract it from the second expression:
\[ \begin{array}{rcccc} & 3a^2 & - & 6 \\ & 2a^2 & + & 1 \\ \text{Change signs:} & (-) & & (-) \\ \hline & a^2 & - & 7 \end{array} \]
In simple words: Subtract the smaller expression from the larger expression to find how much bigger or smaller they are.

Exam Tip: Read the phrasing carefully. "A is how much less than B" always translates to the mathematical operation \( B - A \).

Question 9. If x = 6a + 8b + 9c ; y = 2b - 3a - 6c and z = c - b + 3a ; find
(i) x + y + z
(ii) x - y + z
(iii) 2x - y - 3z
(iv) 3y - 2z - 5x
Answer: By substituting the given values of \( x \), \( y \), and \( z \), we can simplify each expression:
(i) Adding all three expressions:
\[ \begin{array}{rccccc} x & = & 6a & + & 8b & + & 9c \\ y & = & -3a & + & 2b & - & 6c \\ z & = & 3a & - & b & + & c \\ \hline x+y+z & = & 6a & + & 9b & + & 4c \end{array} \]

(ii) For \( x - y + z \):
\( = (6a + 8b + 9c) - (-3a + 2b - 6c) + (3a - b + c) \)
Expand the brackets:
\( = 6a + 8b + 9c + 3a - 2b + 6c + 3a - b + c \)
Combine the like terms:
\( = (6a + 3a + 3a) + (8b - 2b - b) + (9c + 6c + c) \)
\( = 12a + 5b + 16c \)

(iii) For \( 2x - y - 3z \):
\( = 2(6a + 8b + 9c) - (-3a + 2b - 6c) - 3(3a - b + c) \)
Expand the brackets:
\( = 12a + 16b + 18c + 3a - 2b + 6c - 9a + 3b - 3c \)
Combine the like terms:
\( = (12a + 3a - 9a) + (16b - 2b + 3b) + (18c + 6c - 3c) \)
\( = 6a + 17b + 21c \)

(iv) For \( 3y - 2z - 5x \):
\( = 3(-3a + 2b - 6c) - 2(3a - b + c) - 5(6a + 8b + 9c) \)
Expand the brackets:
\( = -9a + 6b - 18c - 6a + 2b - 2c - 30a - 40b - 45c \)
Combine the like terms:
\( = (-9a - 6a - 30a) + (6b + 2b - 40b) + (-18c - 2c - 45c) \)
\( = -45a - 32b - 65c \)
In simple words: Replace the letters x, y, and z with their given expressions. Then, expand any brackets and add the similar terms together.

Exam Tip: When expanding brackets with a negative multiplier outside, remember to multiply both the coefficient and the sign of every term inside.

Question 10. The sides of a triangle are x2 – 3xy + 8, 4x2 + 5xy – 3 and 6 – 3x2 + 4xy. Find its perimeter.
Answer: The perimeter is found by summing the lengths of the three sides:
Perimeter \( = (x^2 - 3xy + 8) + (4x^2 + 5xy - 3) + (6 - 3x^2 + 4xy) \)
Rearrange the terms to group like terms:
\( = (x^2 + 4x^2 - 3x^2) + (-3xy + 5xy + 4xy) + (8 - 3 + 6) \)
\( = 2x^2 + 6xy + 11 \)
In simple words: To find the perimeter of a triangle, simply add the expressions for its three sides together.

Exam Tip: Always state the formula "Perimeter = Sum of all sides" clearly before substituting the algebraic values.

Question 11. The perimeter of a triangle is 8y2 – 9y + 4 and its two sides are 3y2 – 5y and 4y2 + 12. Find its third side.
Answer: To find the length of the third side, we subtract the sum of the two known sides from the total perimeter:
First, find the sum of the two given sides:
Sum of two sides \( = (3y^2 - 5y) + (4y^2 + 12) \)
\( = 7y^2 - 5y + 12 \)
Now, subtract this sum from the perimeter:
Third side \( = (8y^2 - 9y + 4) - (7y^2 - 5y + 12) \)
\( = 8y^2 - 9y + 4 - 7y^2 + 5y - 12 \)
\( = y^2 - 4y - 8 \)
In simple words: Add the two known sides of the triangle first. Then, subtract this sum from the perimeter to get the third side.

Exam Tip: Put brackets around the sum of the two sides when subtracting it from the perimeter to distribute the minus sign correctly.

Question 12. The two adjacent sides of a rectangle are 2x2 – 5xy + 3z2 and 4xy – x2 – z2. Find its perimeter.
Answer: The formula for the perimeter of a rectangle is two times the sum of its adjacent sides:
Sum of adjacent sides \( = (2x^2 - 5xy + 3z^2) + (4xy - x^2 - z^2) \)
\( = (2x^2 - x^2) + (-5xy + 4xy) + (3z^2 - z^2) \)
\( = x^2 - xy + 2z^2 \)
Now, multiply this sum by 2 to get the perimeter:
Perimeter \( = 2(x^2 - xy + 2z^2) \)
\( = 2x^2 - 2xy + 4z^2 \)
In simple words: Add the two adjacent sides of the rectangle together, and then multiply the final sum by 2.

Exam Tip: Simplify the sum inside the brackets first before multiplying each term by 2.

Question 13. What must be subtracted from 19x4 + 2x3 + 30x – 37 to get 8x4 + 22x3 – 7x – 60 ?
Answer: To find what needs to be subtracted, we subtract the desired result from the original expression:
Required expression \( = (19x^4 + 2x^3 + 30x - 37) - (8x^4 + 22x^3 - 7x - 60) \)
Expand the expression:
\( = 19x^4 + 2x^3 + 30x - 37 - 8x^4 - 22x^3 + 7x + 60 \)
Group and combine like terms:
\( = (19x^4 - 8x^4) + (2x^3 - 22x^3) + (30x + 7x) + (-37 + 60) \)
\( = 11x^4 - 20x^3 + 37x + 23 \)
In simple words: To find what to subtract, take the second expression away from the first expression.

Exam Tip: To find what must be subtracted from A to get B, the correct operation is \( A - B \).

Question 14. How much smaller is 15x – 18y + 19z than 22x – 20y – 13z + 26 ?
Answer: We find how much smaller the first expression is by subtracting it from the larger expression:
Difference \( = (22x - 20y - 13z + 26) - (15x - 18y + 19z) \)
\( = 22x - 20y - 13z + 26 - 15x + 18y - 19z \)
Combine the like terms:
\( = (22x - 15x) + (-20y + 18y) + (-13z - 19z) + 26 \)
\( = 7x - 2y - 32z + 26 \)
In simple words: Subtract the first expression from the second one to find the difference between them.

Exam Tip: When subtracting \( 15x - 18y + 19z \), make sure to change the sign of every term to \( -15x + 18y - 19z \).

Question 15. How much bigger is 15x2y2 – 18xy2 – 10x2y than -5x2 + 6x2y – 7xy ?
Answer: We find how much larger the first expression is by subtracting the second expression from it:
Difference \( = (15x^2y^2 - 18xy^2 - 10x^2y) - (-5x^2 + 6x^2y - 7xy) \)
Expand the subtraction:
\( = 15x^2y^2 - 18xy^2 - 10x^2y + 5x^2 - 6x^2y + 7xy \)
Group the like terms (only \( -10x^2y \) and \( -6x^2y \) are like terms):
\( = 15x^2y^2 - 18xy^2 + (-10x^2y - 6x^2y) + 5x^2 + 7xy \)
\( = 15x^2y^2 - 18xy^2 - 16x^2y + 5x^2 + 7xy \)
In simple words: Find the difference by subtracting the second expression from the first expression.

Exam Tip: Only subtract the coefficients of identical like terms - do not combine terms with different powers of \( x \) and \( y \).

Exercise 11(C)

Question 1. Multiply:
(i) \( 8ab^2 \) by \( -4a^3b^4 \)
(ii) \( \frac{2}{3}ab \) by \( -\frac{1}{4}a^2b \)
(iii) \( -5cd^2 \) by \( -5cd^2 \)
(iv) \( 4a \) and \( (6a + 7) \)
(v) \( -8x \) and \( (4 - 2x - x^2) \)
(vi) \( 2a^2 - 5a - 4 \) and \( -3a \)
(vii) \( x + 4 \) by \( x - 5 \)
(viii) \( 5a - 1 \) by \( 7a - 3 \)
(ix) \( 12a + 5b \) by \( 7a - b \)
(x) \( x^2 + x + 1 \) by \( 1 - x \)
(xi) \( 2m^2 - 3m - 1 \) and \( 4m^2 - m - 1 \)
(xii) \( a^2 \), \( ab \) and \( b^2 \)
(xiii) \( abx \), \( -3a^2x \) and \( 7b^2x^3 \)
(xiv) \( -3bx \), \( -5xy \) and \( -7b^3y^2 \)
(xv) \( \left(-\frac{3}{2}x^5y^3\right) \) and \( \left(\frac{4}{9}a^2x^3y\right) \)
(xvi) \( \left(-\frac{2}{3}a^7b^2\right) \) and \( \left(-\frac{9}{4}ab^5\right) \)
(xvii) \( (2a^3 - 3a^2b) \) and \( \left(-\frac{1}{2}ab^2\right) \)
(xviii) \( \left(2x + \frac{1}{2}y\right) \) and \( \left(2x - \frac{1}{2}y\right) \)
Answer:
(i) \( 8ab^2 \times \left(-4a^3b^4\right) = [8 \times (-4)] \times \left(a^1 \times a^3\right) \times \left(b^2 \times b^4\right) \)
\( = -32a^{1+3}b^{2+4} = -32a^4b^6 \)

(ii) \( \left(\frac{2}{3}ab\right) \times \left(-\frac{1}{4}a^2b\right) = \left[\frac{2}{3} \times \left(-\frac{1}{4}\right)\right] \times \left(a^1 \times a^2\right) \times \left(b^1 \times b^1\right) \)
\( = -\frac{1}{6}a^{1+2}b^{1+1} = -\frac{1}{6}a^3b^2 \)

(iii) \( \left(-5cd^2\right) \times \left(-5cd^2\right) = [(-5) \times (-5)] \times \left(c^1 \times c^1\right) \times \left(d^2 \times d^2\right) \)
\( = 25c^{1+1}d^{2+2} = 25c^2d^4 \)

(iv) \( 4a(6a + 7) = (4a \times 6a) + (4a \times 7) \)
\( = 24a^2 + 28a \)

(v) \( -8x\left(4 - 2x - x^2\right) = (-8x \times 4) + (-8x \times -2x) + \left(-8x \times -x^2\right) \)
\( = -32x + 16x^2 + 8x^3 \)

(vi) \( \left(2a^2 - 5a - 4\right)(-3a) = \left(2a^2 \times -3a\right) + (-5a \times -3a) + (-4 \times -3a) \)
\( = -6a^3 + 15a^2 + 12a \)

(vii) \( (x+4)(x-5) = x(x-5) + 4(x-5) \)
\( = x^2 - 5x + 4x - 20 = x^2 - x - 20 \)

(viii) \( (5a-1)(7a-3) = 5a(7a-3) - 1(7a-3) \)
\( = 35a^2 - 15a - 7a + 3 = 35a^2 - 22a + 3 \)

(ix) \( (12a+5b)(7a-b) = 12a(7a-b) + 5b(7a-b) \)
\( = 84a^2 - 12ab + 35ab - 5b^2 = 84a^2 + 23ab - 5b^2 \)

(x) \( \left(x^2+x+1\right)(1-x) = 1\left(x^2+x+1\right) - x\left(x^2+x+1\right) \)
\( = x^2 + x + 1 - x^3 - x^2 - x = 1 - x^3 \)

(xi) \( \left(2m^2 - 3m - 1\right)\left(4m^2 - m - 1\right) = 2m^2\left(4m^2 - m - 1\right) - 3m\left(4m^2 - m - 1\right) - 1\left(4m^2 - m - 1\right) \)
\( = 8m^4 - 2m^3 - 2m^2 - 12m^3 + 3m^2 + 3m - 4m^2 + m + 1 \)
\( = 8m^4 - 14m^3 - 3m^2 + 4m + 1 \)

(xii) \( a^2 \times ab \times b^2 = \left(a^2 \times a^1\right) \times \left(b^1 \times b^2\right) \)
\( = a^{2+1}b^{1+2} = a^3b^3 \)

(xiii) \( abx \times \left(-3a^2x\right) \times 7b^2x^3 = [1 \times (-3) \times 7] \times \left(a^1 \times a^2\right) \times \left(b^1 \times b^2\right) \times \left(x^1 \times x^1 \times x^3\right) \)
\( = -21a^{1+2}b^{1+2}x^{1+1+3} = -21a^3b^3x^5 \)

(xiv) \( -3bx \times -5xy \times -7b^3y^2 = [(-3) \times (-5) \times (-7)] \times \left(b^1 \times b^3\right) \times \left(x^1 \times x^1\right) \times \left(y^1 \times y^2\right) \)
\( = -105b^{1+3}x^{1+1}y^{1+2} = -105b^4x^2y^3 \)

(xv) \( \left(-\frac{3}{2}x^5y^3\right) \times \left(\frac{4}{9}a^2x^3y\right) = \left[-\frac{3}{2} \times \frac{4}{9}\right] \times a^2 \times \left(x^5 \times x^3\right) \times \left(y^3 \times y^1\right) \)
\( = -\frac{2}{3}a^2x^{5+3}y^{3+1} = -\frac{2}{3}a^2x^8y^4 \)

(xvi) \( \left(-\frac{2}{3}a^7b^2\right) \times \left(-\frac{9}{4}ab^5\right) = \left[-\frac{2}{3} \times \left(-\frac{9}{4}\right)\right] \times \left(a^7 \times a^1\right) \times \left(b^2 \times b^5\right) \)
\( = \frac{3}{2}a^{7+1}b^{2+5} = \frac{3}{2}a^8b^7 \)

(xvii) \( \left(2a^3 - 3a^2b\right) \times \left(-\frac{1}{2}ab^2\right) = \left(2a^3 \times -\frac{1}{2}ab^2\right) - \left(3a^2b \times -\frac{1}{2}ab^2\right) \)
\( = -a^4b^2 + \frac{3}{2}a^3b^3 \)

(xviii) \( \left(2x + \frac{1}{2}y\right)\left(2x - \frac{1}{2}y\right) = (2x)^2 - \left(\frac{1}{2}y\right)^2 \)
\( = 4x^2 - \frac{1}{4}y^2 \)

In simple words: To multiply algebraic terms, first multiply the numbers together. Next, multiply the same variables by adding their power values.

Exam Tip: Be very careful with positive and negative signs when multiplying terms, and remember to add exponents of identical bases.

Question 2. Multiply:
(i) \( -5x^2 - 8xy + 6y^2 - 3 \) by \( -3xy \)
(ii) \( 3 - \frac{2}{3}xy + \frac{5}{7}xy^2 - \frac{16}{21}x^2y \) by \( -21x^2y^2 \)
(iii) \( 6x^3 - 5x + 10 \) by \( 4 - 3x^2 \)
(iv) \( 2y - 4y^3 + 6y^5 \) by \( y^2 + y - 3 \)
(v) \( 5p^2 + 25pq + 4q^2 \) by \( 2p^2 - 2pq + 3q^2 \)
Answer:
(i) Multiply each term of \( \left(-5x^2 - 8xy + 6y^2 - 3\right) \) by \( -3xy \):
\( = \left(-5x^2\right) \times (-3xy) - (8xy) \times (-3xy) + \left(6y^2\right) \times (-3xy) - (3) \times (-3xy) \)
\( = 15x^3y + 24x^2y^2 - 18xy^3 + 9xy \)

(ii) Multiply each term by \( -21x^2y^2 \):
\( = 3 \times \left(-21x^2y^2\right) - \left(\frac{2}{3}xy\right) \times \left(-21x^2y^2\right) + \left(\frac{5}{7}xy^2\right) \times \left(-21x^2y^2\right) - \left(\frac{16}{21}x^2y\right) \times \left(-21x^2y^2\right) \)
\( = -63x^2y^2 + 14x^3y^3 - 15x^3y^4 + 16x^4y^3 \)

(iii) We can multiply these two expressions term-by-term:
\( \left(6x^3 - 5x + 10\right)\left(4 - 3x^2\right) \)
\( = 4\left(6x^3 - 5x + 10\right) - 3x^2\left(6x^3 - 5x + 10\right) \)
\( = \left(24x^3 - 20x + 40\right) - \left(18x^5 - 15x^3 + 30x^2\right) \)
\( = -18x^5 + 15x^3 + 24x^3 - 30x^2 - 20x + 40 \)
\( = -18x^5 + 39x^3 - 30x^2 - 20x + 40 \)

(iv) Multiply term-by-term:
\( \left(2y - 4y^3 + 6y^5\right)\left(y^2 + y - 3\right) \)
\( = y^2\left(2y - 4y^3 + 6y^5\right) + y\left(2y - 4y^3 + 6y^5\right) - 3\left(2y - 4y^3 + 6y^5\right) \)
\( = \left(2y^3 - 4y^5 + 6y^7\right) + \left(2y^2 - 4y^4 + 6y^6\right) - \left(6y - 12y^3 + 18y^5\right) \)
\( = 6y^7 + 6y^6 + \left(-4y^5 - 18y^5\right) - 4y^4 + \left(2y^3 + 12y^3\right) + 2y^2 - 6y \)
\( = 6y^7 + 6y^6 - 22y^5 - 4y^4 + 14y^3 + 2y^2 - 6y \)

(v) Multiply term-by-term:
\( \left(5p^2 + 25pq + 4q^2\right)\left(2p^2 - 2pq + 3q^2\right) \)
\( = 2p^2\left(5p^2 + 25pq + 4q^2\right) - 2pq\left(5p^2 + 25pq + 4q^2\right) + 3q^2\left(5p^2 + 25pq + 4q^2\right) \)
\( = \left(10p^4 + 50p^3q + 8p^2q^2\right) - \left(10p^3q + 50p^2q^2 + 8pq^3\right) + \left(15p^2q^2 + 75pq^3 + 12q^4\right) \)
\( = 10p^4 + (50 - 10)p^3q + (8 - 50 + 15)p^2q^2 + (-8 + 75)pq^3 + 12q^4 \)
\( = 10p^4 + 40p^3q - 27p^2q^2 + 67pq^3 + 12q^4 \)

In simple words: Multiplying longer expressions is like distributing. Multiply each term in the first bracket by each term in the second, then add similar terms together.

Exam Tip: Keep track of each term as you multiply, and double-check your addition for any terms that have the same variables and powers.

Question 3. Simplify:
(i) \( (7x - 8)(3x + 2) \)
(ii) \( (px - q)(px + q) \)
(iii) \( (5a + 5b - c)(2b - 3c) \)
(iv) \( (4x - 5y)(5x - 4y) \)
(v) \( (3y + 4z)(3y - 4z) + (2y + 7z)(y + z) \)
Answer:
(i) \( (7x - 8)(3x + 2) = 7x(3x + 2) - 8(3x + 2) \)
\( = 21x^2 + 14x - 24x - 16 \)
\( = 21x^2 - 10x - 16 \)

(ii) \( (px - q)(px + q) = (px)^2 - q^2 \)
\( = p^2x^2 - q^2 \)

(iii) \( (5a + 5b - c)(2b - 3c) = 5a(2b - 3c) + 5b(2b - 3c) - c(2b - 3c) \)
\( = 10ab - 15ac + 10b^2 - 15bc - 2bc + 3c^2 \)
\( = 10ab + 10b^2 + 3c^2 - 17bc - 15ac \)

(iv) \( (4x - 5y)(5x - 4y) = 4x(5x - 4y) - 5y(5x - 4y) \)
\( = 20x^2 - 16xy - 25xy + 20y^2 \)
\( = 20x^2 - 41xy + 20y^2 \)

(v) \( (3y + 4z)(3y - 4z) + (2y + 7z)(y + z) \)
First, simplify the first part using the difference of squares:
\( (3y + 4z)(3y - 4z) = 9y^2 - 16z^2 \)
Next, multiply the second part term-by-term:
\( (2y + 7z)(y + z) = 2y(y + z) + 7z(y + z) = 2y^2 + 2yz + 7yz + 7z^2 = 2y^2 + 9yz + 7z^2 \)
Now, combine both simplified parts:
\( = \left(9y^2 - 16z^2\right) + \left(2y^2 + 9yz + 7z^2\right) \)
\( = 11y^2 + 9yz - 9z^2 \)

In simple words: Break down the problem by multiplying the brackets first. After that, group and add the matching variable terms.

Exam Tip: You can save time by using special identities like \( (a-b)(a+b) = a^2 - b^2 \) for sub-parts like (ii) and (v).

Question 4. The adjacent sides of a rectangle are \( x^2 - 4xy + 7y^2 \) and \( x^3 - 5xy^2 \). Find its area.
Answer:
The side lengths of the rectangle are given as \( x^2 - 4xy + 7y^2 \) and \( x^3 - 5xy^2 \).
The area of a rectangle is calculated by multiplying its two adjacent sides:
\( \text{Area} = \left(x^2 - 4xy + 7y^2\right)\left(x^3 - 5xy^2\right) \)
\( = x^2\left(x^3 - 5xy^2\right) - 4xy\left(x^3 - 5xy^2\right) + 7y^2\left(x^3 - 5xy^2\right) \)
\( = x^5 - 5x^3y^2 - 4x^4y + 20x^2y^3 + 7x^3y^2 - 35xy^4 \)
Now, group the similar terms:
\( = x^5 - 4x^4y + (-5 + 7)x^3y^2 + 20x^2y^3 - 35xy^4 \)
\( = x^5 - 4x^4y + 2x^3y^2 + 20x^2y^3 - 35xy^4 \text{ sq. units} \)

In simple words: To find the area of a rectangle, multiply the two given side lengths together.

Exam Tip: Do not forget to include "sq. units" or "square units" in your final answer for area problems.

Question 5. The base and the altitude of a triangle are \( (3x - 4y) \) and \( (6x + 5y) \) respectively. Find its area.
Answer:
The area of a triangle is given by the formula:
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{altitude} \)
Substituting the given base and altitude:
\( \text{Area} = \frac{1}{2}(3x - 4y)(6x + 5y) \)
\( = \frac{1}{2}[3x(6x + 5y) - 4y(6x + 5y)] \)
\( = \frac{1}{2}\left[18x^2 + 15xy - 24xy - 20y^2\right] \)
\( = \frac{1}{2}\left(18x^2 - 9xy - 20y^2\right) \text{ sq. units} \)

In simple words: Multiply the base by the height, and then divide the entire result by 2 to get the triangle's area.

Exam Tip: Be careful to apply the factor of \( \frac{1}{2} \) to the entire polynomial product, not just the first term.

Question 6. Multiply \( -4xy^3 \) and \( 6x^2y \) and verify your result for \( x = 2 \) and \( y = 1 \).
Answer:
First, let us find the product of the two given terms:
\( \left(-4xy^3\right) \times \left(6x^2y\right) = [-4 \times 6] \times \left(x^1 \times x^2\right) \times \left(y^3 \times y^1\right) \)
\( = -24x^{1+2}y^{3+1} = -24x^3y^4 \)

Now, let us verify this product using \( x = 2 \) and \( y = 1 \):
Left Hand Side (L.H.S.):
\( \text{L.H.S.} = \left(-4xy^3\right) \times \left(6x^2y\right) \)
\( = \left[-4(2)(1)^3\right] \times \left[6(2)^2(1)\right] \)
\( = (-8) \times (6 \times 4 \times 1) \)
\( = -8 \times 24 = -192 \)

Right Hand Side (R.H.S.):
\( \text{R.H.S.} = -24x^3y^4 \)
\( = -24(2)^3(1)^4 \)
\( = -24 \times 8 \times 1 = -192 \)

Since the Left Hand Side is equal to the Right Hand Side (\( \text{L.H.S.} = \text{R.H.S.} = -192 \)), the result is verified.

In simple words: Multiply the terms first to get an answer. Then put the numbers in both the start question and your answer to check if they match.

Exam Tip: Clearly show the calculations for both L.H.S. and R.H.S. to get full credit for verification questions.

Question 7. Find the value of \( \left(3x^3\right) \times \left(-5xy^2\right) \times \left(2x^2yz^3\right) \) for \( x = 1 \), \( y = 2 \) and \( z = 3 \).
Answer:
We can substitute the values \( x = 1 \), \( y = 2 \), and \( z = 3 \) directly into the expression:
\( \left(3x^3\right) \times \left(-5xy^2\right) \times \left(2x^2yz^3\right) \)
\( = \left[3(1)^3\right] \times \left[-5(1)(2)^2\right] \times \left[2(1)^2(2)(3)^3\right] \)
\( = (3 \times 1) \times (-5 \times 1 \times 4) \times (2 \times 1 \times 2 \times 27) \)
\( = 3 \times (-20) \times (108) \)
\( = -60 \times 108 = -6480 \)

In simple words: Put the given numbers in place of the letters and calculate the final multiplication.

Exam Tip: Work out the powers of each number before multiplying them with their coefficients to avoid calculation mistakes.

Question 8. Evaluate \( \left(3x^4y^2\right) \times \left(2x^2y^3\right) \) for \( x = 1 \) and \( y = 2 \).
Answer:
Substitute \( x = 1 \) and \( y = 2 \) into the expression:
\( \left(3x^4y^2\right) \times \left(2x^2y^3\right) \)
\( = \left[3(1)^4(2)^2\right] \times \left[2(1)^2(2)^3\right] \)
\( = (3 \times 1 \times 4) \times (2 \times 1 \times 8) \)
\( = 12 \times 16 = 192 \)

In simple words: Replace the letters with the given values and then do the math.

Exam Tip: Be sure to raise the numbers to their correct powers first before doing the multiplications.

Question 9. Evaluate \( \left(x^5\right) \times \left(3x^2\right) \times (-2x) \) for \( x = 1 \).
Answer:
We can substitute \( x = 1 \) into the given expression:
\( \left(x^5\right) \times \left(3x^2\right) \times (-2x) \)
\( = \left(1^5\right) \times \left[3(1)^2\right] \times [-2(1)] \)
\( = 1 \times (3 \times 1) \times (-2) \)
\( = 1 \times 3 \times (-2) = -6 \)

In simple words: Change the letter \( x \) to \( 1 \) and multiply the numbers together.

Exam Tip: Multiplying any power of 1 is just 1, which makes this evaluation very quick and simple.

Question 10. If \( x = 2 \) and \( y = 1 \); find the value of \( \left(-4x^2y^3\right) \times \left(-5x^2y^5\right) \).
Answer:
Substitute the values \( x = 2 \) and \( y = 1 \) into the given expression:
\( \left(-4x^2y^3\right) \times \left(-5x^2y^5\right) \)
\( = \left[-4(2)^2(1)^3\right] \times \left[-5(2)^2(1)^5\right] \)
\( = (-4 \times 4 \times 1) \times (-5 \times 4 \times 1) \)
\( = (-16) \times (-20) = 320 \)

In simple words: Put \( 2 \) instead of \( x \) and \( 1 \) instead of \( y \), then multiply the parts to find the final number.

Exam Tip: Multiplying two negative values results in a positive final answer.

Question 11. Evaluate:
(i) \( (3x - 2)(x + 5) \) for \( x = 2 \).
(ii) \( (2x - 5y)(2x + 3y) \) for \( x = 2 \) and \( y = 3 \).
(iii) \( xz\left(x^2 + y^2\right) \) for \( x = 2 \), \( y = 1 \) and \( z = 1 \).
Answer:
(i) For \( x = 2 \):
\( (3x - 2)(x + 5) = (3 \times 2 - 2)(2 + 5) \)
\( = (6 - 2) \times 7 = 4 \times 7 = 28 \)

(ii) For \( x = 2 \) and \( y = 3 \):
\( (2x - 5y)(2x + 3y) = [2(2) - 5(3)] \times [2(2) + 3(3)] \)
\( = (4 - 15) \times (4 + 9) \)
\( = -11 \times 13 = -143 \)

(iii) For \( x = 2 \), \( y = 1 \), and \( z = 1 \):
\( xz\left(x^2 + y^2\right) = (2)(1) \times \left(2^2 + 1^2\right) \)
\( = 2 \times (4 + 1) = 2 \times 5 = 10 \)

In simple words: Replace the letters with the given numbers in each part and compute the result step-by-step.

Exam Tip: Be sure to compute terms inside brackets first, following the standard BODMAS rules.

Question 12. Evaluate:
(i) \( x(x - 5) + 2 \) for \( x = 1 \).
(ii) \( xy^2(x - 5y) + 1 \) for \( x = 2 \) and \( y = 1 \).
(iii) \( 2x(3x - 5) - 5(x - 2) - 18 \) for \( x = 2 \).
Answer:
(i) For \( x = 1 \):
\( x(x - 5) + 2 = 1(1 - 5) + 2 \)
\( = 1(-4) + 2 = -4 + 2 = -2 \)

(ii) For \( x = 2 \) and \( y = 1 \):
\( xy^2(x - 5y) + 1 = (2)(1)^2 \times (2 - 5 \times 1) + 1 \)
\( = 2(1) \times (2 - 5) + 1 \)
\( = 2 \times (-3) + 1 = -6 + 1 = -5 \)

(iii) For \( x = 2 \):
\( 2x(3x - 5) - 5(x - 2) - 18 = 2(2)(3 \times 2 - 5) - 5(2 - 2) - 18 \)
\( = 4(6 - 5) - 5(0) - 18 \)
\( = 4(1) - 0 - 18 = 4 - 18 = -14 \)

In simple words: Substitute the given values of the variables into the expressions and simplify them.

Exam Tip: Be cautious when subtracting larger numbers from smaller ones to avoid making sign errors.

Question 13. Multiply and then verify : \( -3x^2y^2 \) and \( (x - 2y) \) for \( x = 1 \) and \( y = 2 \).
Answer:
First, find the product of the given terms:
\( -3x^2y^2 \times (x - 2y) = \left(-3x^2y^2 \times x\right) - \left(-3x^2y^2 \times 2y\right) \)
\( = -3x^3y^2 + 6x^2y^3 \)

Now, verify this result using \( x = 1 \) and \( y = 2 \):
Left Hand Side (L.H.S.):
\( \text{L.H.S.} = -3x^2y^2 \times (x - 2y) \)
\( = -3(1)^2(2)^2 \times (1 - 2 \times 2) \)
\( = -3(1)(4) \times (1 - 4) \)
\( = -12 \times (-3) = 36 \)

Right Hand Side (R.H.S.):
\( \text{R.H.S.} = -3x^3y^2 + 6x^2y^3 \)
\( = -3(1)^3(2)^2 + 6(1)^2(2)^3 \)
\( = -3(1)(4) + 6(1)(8) \)
\( = -12 + 48 = 36 \)

Since the Left Hand Side is equal to the Right Hand Side (\( \text{L.H.S.} = \text{R.H.S.} = 36 \)), the result is verified.

In simple words: Multiply the terms first to get an algebraic answer. Then check by substituting the values into both forms to see if the numerical results are identical.

Exam Tip: Make sure to perform exponentiation before multiplication in both the L.H.S. and R.H.S. calculations.

Question 14. Multiply:
(i) \( 2x^2 - 4x + 5 \) by \( x^2 + 3x - 7 \)
(ii) \( (ab - 1)(3 - 2ab) \)
Answer:
(i) Multiply each term of the first expression by the entire second expression:
\( \left(2x^2 - 4x + 5\right)\left(x^2 + 3x - 7\right) = 2x^2\left(x^2 + 3x - 7\right) - 4x\left(x^2 + 3x - 7\right) + 5\left(x^2 + 3x - 7\right) \)
\( = \left(2x^4 + 6x^3 - 14x^2\right) - \left(4x^3 + 12x^2 - 28x\right) + \left(5x^2 + 15x - 35\right) \)
\( = 2x^4 + 6x^3 - 14x^2 - 4x^3 - 12x^2 + 28x + 5x^2 + 15x - 35 \)
Group the similar terms together:
\( = 2x^4 + (6 - 4)x^3 + (-14 - 12 + 5)x^2 + (28 + 15)x - 35 \)
\( = 2x^4 + 2x^3 - 21x^2 + 43x - 35 \)

(ii) Multiply the two binomials term-by-term:
\( (ab - 1)(3 - 2ab) = ab(3 - 2ab) - 1(3 - 2ab) \)
\( = 3ab - 2a^2b^2 - 3 + 2ab \)
Group the like terms:
\( = -2a^2b^2 + (3 + 2)ab - 3 \)
\( = -2a^2b^2 + 5ab - 3 \)

In simple words: Multiply each part of the first expression by every part of the second expression, then group and combine the matching variables.

Exam Tip: Be careful with sign distribution when expanding, especially when multiplying by a negative term like \( -4x \) or \( -1 \).

Question 15. Simplify: \( (5 - x)(6 - 5x)(2 - x) \).
Answer:
First, find the product of the first two brackets:
\( (5 - x)(6 - 5x) = 5(6 - 5x) - x(6 - 5x) \)
\( = 30 - 25x - 6x + 5x^2 \)
\( = 5x^2 - 31x + 30 \)

Now, multiply this resulting trinomial by the third bracket \( (2 - x) \):
\( \left(5x^2 - 31x + 30\right)(2 - x) = 2\left(5x^2 - 31x + 30\right) - x\left(5x^2 - 31x + 30\right) \)
\( = \left(10x^2 - 62x + 60\right) - \left(5x^3 - 31x^2 + 30x\right) \)
\( = 10x^2 - 62x + 60 - 5x^3 + 31x^2 - 30x \)
Group and combine the like terms:
\( = -5x^3 + (10 + 31)x^2 + (-62 - 30)x + 60 \)
\( = -5x^3 + 41x^2 - 92x + 60 \)

In simple words: Multiply two brackets first to get a single larger expression. Then, multiply that new expression by the last bracket.

Exam Tip: When simplifying a product of three binomials, multiply them two at a time and write down each intermediate step clearly to prevent sign or exponent mistakes.

Exercise 11(D)

Question 1. Divide:
(i) \( -70a^3 \) by \( 14a^2 \)
(ii) \( 24x^3y^3 \) by \( -8y^2 \)
(iii) \( 15a^4b \) by \( -5a^3b \)
(iv) \( -24x^4d^3 \) by \( -2x^2d^5 \)
(v) \( 63a^4b^5c^6 \) by \( -9a^2b^4c^3 \)
(vi) \( 8x - 10y + 6c \) by \( 2 \)
(vii) \( 15a^3b^4 - 10a^4b^3 - 25a^3b^6 \) by \( -5a^3b^2 \)
(viii) \( -14x^6y^3 - 21x^4y^5 + 7x^5y^4 \) by \( 7x^2y^2 \)
(ix) \( a^2 + 7a + 12 \) by \( a + 4 \)
(x) \( x^2 + 3x - 54 \) by \( x - 6 \)
(xi) \( 12x^2 + 7xy - 12y^2 \) by \( 3x + 4y \)
(xii) \( x^6 - 8 \) by \( x^2 - 2 \)
(xiii) \( 6x^3 - 13x^2 - 13x + 30 \) by \( 2x^2 - x - 6 \)
(xiv) \( 4a^2 + 12ab + 9b^2 - 25c^2 \) by \( 2a + 3b + 5c \)
(xv) \( 16 + 8x + x^6 - 8x^3 - 2x^4 + x^2 \) by \( x + 4 - x^3 \)
Answer:
(i) \( \frac{-70a^3}{14a^2} = \left(\frac{-70}{14}\right) \left(\frac{a^3}{a^2}\right) = -5 \cdot a^{3-2} = -5a \)

(ii) \( \frac{24x^3y^3}{-8y^2} = \left(\frac{24}{-8}\right) (x^3) \left(\frac{y^3}{y^2}\right) = -3 \cdot x^3 \cdot y^{3-2} = -3x^3y \)

(iii) \( \frac{15a^4b}{-5a^3b} = \left(\frac{15}{-5}\right) \left(\frac{a^4}{a^3}\right) \left(\frac{b}{b}\right) = -3 \cdot a^{4-3} \cdot b^{1-1} = -3 \cdot a \cdot 1 = -3a \)

(iv) \( \frac{-24x^4d^3}{-2x^2d^5} = \left(\frac{-24}{-2}\right) \left(\frac{x^4}{x^2}\right) \left(\frac{d^3}{d^5}\right) = 12 \cdot x^{4-2} \cdot d^{3-5} = 12 \cdot x^2 \cdot d^{-2} = \frac{12x^2}{d^2} \)

(v) \( \frac{63a^4b^5c^6}{-9a^2b^4c^3} = \left(\frac{63}{-9}\right) \left(\frac{a^4}{a^2}\right) \left(\frac{b^5}{b^4}\right) \left(\frac{c^6}{c^3}\right) = -7 \cdot a^{4-2} \cdot b^{5-4} \cdot c^{6-3} = -7a^2bc^3 \)

(vi) \( \frac{8x - 10y + 6c}{2} = \frac{8x}{2} - \frac{10y}{2} + \frac{6c}{2} = 4x - 5y + 3c \)

(vii) \( \frac{15a^3b^4 - 10a^4b^3 - 25a^3b^6}{-5a^3b^2} = \frac{15a^3b^4}{-5a^3b^2} - \frac{10a^4b^3}{-5a^3b^2} - \frac{25a^3b^6}{-5a^3b^2} = -3b^{4-2} + 2a^{4-3}b^{3-2} + 5b^{6-2} = -3b^2 + 2ab + 5b^4 \)

(viii) \( \frac{-14x^6y^3 - 21x^4y^5 + 7x^5y^4}{7x^2y^2} = \frac{-14x^6y^3}{7x^2y^2} - \frac{21x^4y^5}{7x^2y^2} + \frac{7x^5y^4}{7x^2y^2} = -2x^{6-2}y^{3-2} - 3x^{4-2}y^{5-2} + x^{5-2}y^{4-2} = -2x^4y - 3x^2y^3 + x^3y^2 \)

(ix) To divide \( a^2 + 7a + 12 \) by \( a + 4 \), we use polynomial long division: \[ \begin{array}{rll} a + 3 && \\ a + 4 \ \overline{\smash{)} \ a^2 + 7a + 12} && \\ \underline{a^2 + 4a} && \\ 3a + 12 && \\ \underline{3a + 12} && \\ 0 && \end{array} \] Hence, the quotient is \( a + 3 \).

(x) To divide \( x^2 + 3x - 54 \) by \( x - 6 \): \[ \begin{array}{rll} x + 9 && \\ x - 6 \ \overline{\smash{)} \ x^2 + 3x - 54} && \\ \underline{x^2 - 6x} && \\ 9x - 54 && \\ \underline{9x - 54} && \\ 0 && \end{array} \] Hence, the quotient is \( x + 9 \).

(xi) To divide \( 12x^2 + 7xy - 12y^2 \) by \( 3x + 4y \): \[ \begin{array}{rll} 4x - 3y && \\ 3x + 4y \ \overline{\smash{)} \ 12x^2 + 7xy - 12y^2} && \\ \underline{12x^2 + 16xy} && \\ -9xy - 12y^2 && \\ \underline{-9xy - 12y^2} && \\ 0 && \end{array} \] Hence, the quotient is \( 4x - 3y \).

(xii) To divide \( x^6 - 8 \) by \( x^2 - 2 \): \[ \begin{array}{rll} x^4 + 2x^2 + 4 && \\ x^2 - 2 \ \overline{\smash{)} \ x^6 + 0x^4 + 0x^2 - 8} && \\ \underline{x^6 - 2x^4} && \\ 2x^4 + 0x^2 - 8 && \\ \underline{2x^4 - 4x^2} && \\ 4x^2 - 8 && \\ \underline{4x^2 - 8} && \\ 0 && \end{array} \] Hence, the quotient is \( x^4 + 2x^2 + 4 \).

(xiii) To divide \( 6x^3 - 13x^2 - 13x + 30 \) by \( 2x^2 - x - 6 \): \[ \begin{array}{rll} 3x - 5 && \\ 2x^2 - x - 6 \ \overline{\smash{)} \ 6x^3 - 13x^2 - 13x + 30} && \\ \underline{6x^3 - 3x^2 - 18x} && \\ -10x^2 + 5x + 30 && \\ \underline{-10x^2 + 5x + 30} && \\ 0 && \end{array} \] Hence, the quotient is \( 3x - 5 \).

(xiv) To divide \( 4a^2 + 12ab + 9b^2 - 25c^2 \) by \( 2a + 3b + 5c \): \[ \begin{array}{rll} 2a + 3b - 5c && \\ 2a + 3b + 5c \ \overline{\smash{)} \ 4a^2 + 12ab + 9b^2 - 25c^2} && \\ \underline{4a^2 + 6ab \phantom{{} + 9b^2 - 25c^2} + 10ca} && \\ 6ab + 9b^2 - 25c^2 - 10ca && \\ \underline{6ab + 9b^2 \phantom{{} - 25c^2 - 10ca} + 15bc} && \\ -10ca - 25c^2 - 15bc && \\ \underline{-10ca - 25c^2 - 15bc} && \\ 0 && \end{array} \] Hence, the quotient is \( 2a + 3b - 5c \).

(xv) Arrange terms in descending powers of \( x \). This yields the dividend \( x^6 - 2x^4 - 8x^3 + x^2 + 8x + 16 \) and divisor \( -x^3 + x + 4 \): \[ \begin{array}{rll} -x^3 + x + 4 && \\ -x^3 + x + 4 \ \overline{\smash{)} \ x^6 - 2x^4 - 8x^3 + x^2 + 8x + 16} && \\ \underline{x^6 - x^4 - 4x^3} && \\ -x^4 - 4x^3 + x^2 + 8x + 16 && \\ \underline{-x^4 + x^2 + 4x} && \\ -4x^3 + 4x + 16 && \\ \underline{-4x^3 + 4x + 16} && \\ 0 && \end{array} \] Hence, the quotient is \( -x^3 + x + 4 \).
In simple words: To divide algebraic terms, divide the numerical coefficients first. Then, subtract the exponent of the variable in the denominator from the exponent of the same variable in the numerator. For larger polynomials, use long division just like you do with normal numbers.

Exam Tip: Pay close attention to negative signs during division. If there is a single negative sign in either the numerator or denominator, the quotient will be negative.

Question 2. Find the quotient and the remainder (if any) when :
(i) \( a^3 - 5a^2 + 8a + 15 \) is divided by \( a + 1 \).
(ii) \( 3x^4 + 6x^3 - 6x^2 + 2x - 7 \) is divided by \( x - 3 \).
(iii) \( 6x^2 + x - 15 \) is divided by \( 3x + 5 \).
(iv) \( 6y^5 + 30y^4 + 18y^3 + 6y^2 + 15y + 3 \) is divided by \( 2y^3 + 1 \).
In each case, verify your answer.
Answer:
(i) Dividing \( a^3 - 5a^2 + 8a + 15 \) by \( a + 1 \): \[ \begin{array}{rll} a^2 - 6a + 14 && \\ a + 1 \ \overline{\smash{)} \ a^3 - 5a^2 + 8a + 15} && \\ \underline{a^3 + a^2} && \\ -6a^2 + 8a + 15 && \\ \underline{-6a^2 - 6a} && \\ 14a + 15 && \\ \underline{14a + 14} && \\ 1 && \end{array} \] So, Quotient = \( a^2 - 6a + 14 \) and Remainder = \( 1 \).
Verification:
We know that: \( \text{Dividend} = \text{Quotient} \times \text{Divisor} + \text{Remainder} \)
\( = (a^2 - 6a + 14)(a + 1) + 1 \)
\( = a(a^2 - 6a + 14) + 1(a^2 - 6a + 14) + 1 \)
\( = a^3 - 6a^2 + 14a + a^2 - 6a + 14 + 1 \)
\( = a^3 - 5a^2 + 8a + 15 \)
Since this matches the original dividend, our result is verified.

(ii) Dividing \( 3x^4 + 6x^3 - 6x^2 + 2x - 7 \) by \( x - 3 \): \[ \begin{array}{rll} 3x^3 + 15x^2 + 39x + 119 && \\ x - 3 \ \overline{\smash{)} \ 3x^4 + 6x^3 - 6x^2 + 2x - 7} && \\ \underline{3x^4 - 9x^3} && \\ 15x^3 - 6x^2 + 2x - 7 && \\ \underline{15x^3 - 45x^2} && \\ 39x^2 + 2x - 7 && \\ \underline{39x^2 - 117x} && \\ 119x - 7 && \\ \underline{119x - 357} && \\ 350 && \end{array} \] So, Quotient = \( 3x^3 + 15x^2 + 39x + 119 \) and Remainder = \( 350 \).
Verification:
Using the relation: \( \text{Dividend} = \text{Quotient} \times \text{Divisor} + \text{Remainder} \)
\( = (3x^3 + 15x^2 + 39x + 119)(x - 3) + 350 \)
\( = x(3x^3 + 15x^2 + 39x + 119) - 3(3x^3 + 15x^2 + 39x + 119) + 350 \)
\( = 3x^4 + 15x^3 + 39x^2 + 119x - 9x^3 - 45x^2 - 117x - 357 + 350 \)
\( = 3x^4 + 6x^3 - 6x^2 + 2x - 7 \)
Since this matches our initial polynomial, our calculation is correct.

(iii) Dividing \( 6x^2 + x - 15 \) by \( 3x + 5 \): \[ \begin{array}{rll} 2x - 3 && \\ 3x + 5 \ \overline{\smash{)} \ 6x^2 + x - 15} && \\ \underline{6x^2 + 10x} && \\ -9x - 15 && \\ \underline{-9x - 15} && \\ 0 && \end{array} \] So, Quotient = \( 2x - 3 \) and Remainder = \( 0 \).
Verification:
Using the relation: \( \text{Dividend} = \text{Quotient} \times \text{Divisor} + \text{Remainder} \)
\( = (2x - 3)(3x + 5) + 0 \)
\( = 2x(3x + 5) - 3(3x + 5) \)
\( = 6x^2 + 10x - 9x - 15 \)
\( = 6x^2 + x - 15 \)
This is identical to our original dividend, which confirms our division is right.

(iv) Dividing \( 6y^5 + 30y^4 + 18y^3 + 6y^2 + 15y + 3 \) by \( 2y^3 + 1 \): \[ \begin{array}{rll} 3y^2 + 15y + 9 && \\ 2y^3 + 1 \ \overline{\smash{)} \ 6y^5 + 30y^4 + 18y^3 + 6y^2 + 15y + 3} && \\ \underline{6y^5 \phantom{{} + 30y^4 + 18y^3} + 3y^2} && \\ 30y^4 + 18y^3 + 3y^2 + 15y + 3 && \\ \underline{30y^4 \phantom{{} + 18y^3 + 3y^2} + 15y} && \\ 18y^3 + 3y^2 + 3 && \\ \underline{18y^3 \phantom{{} + 3y^2} + 9} && \\ 3y^2 - 6 && \end{array} \] So, Quotient = \( 3y^2 + 15y + 9 \) and Remainder = \( 3y^2 - 6 \).
Verification:
Using the relation: \( \text{Dividend} = \text{Quotient} \times \text{Divisor} + \text{Remainder} \)
\( = (3y^2 + 15y + 9)(2y^3 + 1) + (3y^2 - 6) \)
\( = 2y^3(3y^2 + 15y + 9) + 1(3y^2 + 15y + 9) + 3y^2 - 6 \)
\( = 6y^5 + 30y^4 + 18y^3 + 3y^2 + 15y + 9 + 3y^2 - 6 \)
\( = 6y^5 + 30y^4 + 18y^3 + (3y^2 + 3y^2) + 15y + (9 - 6) \)
\( = 6y^5 + 30y^4 + 18y^3 + 6y^2 + 15y + 3 \)
The result matches the original expression, proving the answer is accurate.
In simple words: To check your polynomial division, multiply the quotient by the divisor and add the remainder. If you get the exact starting polynomial back, your division is correct!

Exam Tip: Keep powers of variables aligned in columns when doing long division. If a term is missing (like no \( y \) term), leave a blank space or use a zero placeholder to avoid mistakes.

Question 3. The area of a rectangle is \( x^3 - 8x^2 + 7 \) and one of its sides is \( x - 1 \). Find the length of the adjacent side.
Answer:
Given:
Area of the rectangle = \( x^3 - 8x^2 + 7 \)
Length of one side = \( x - 1 \)
Since the area of a rectangle is the product of its adjacent sides, we find the second side by dividing the area by the first side:
\( \text{Length of adjacent side} = \frac{\text{Area of rectangle}}{\text{Length of one side}} = \frac{x^3 - 8x^2 + 7}{x - 1} \)
We perform polynomial long division of \( x^3 - 8x^2 + 7 \) by \( x - 1 \), adding a placeholder term \( 0x \) for the missing linear term:
\[ \begin{array}{rll} x^2 - 7x - 7 && \\ x - 1 \ \overline{\smash{)} \ x^3 - 8x^2 + 0x + 7} && \\ \underline{x^3 - x^2} && \\ -7x^2 + 0x + 7 && \\ \underline{-7x^2 + 7x} && \\ -7x + 7 && \\ \underline{-7x + 7} && \\ 0 && \end{array} \] So, the adjacent side is \( x^2 - 7x - 7 \).
In simple words: Since multiplying the two sides of a rectangle gives its area, dividing the area by one side will give you the other side.

Exam Tip: Be sure to write \( 0x \) as a placeholder for the missing \( x \) term in the dividend to keep your subtraction steps organized.

Question 4. The product of two numbers is \( 16x^4 - 1 \). If one number is \( 2x - 1 \), find the other.
Answer:
Given:
Product of the two numbers = \( 16x^4 - 1 \)
First number = \( 2x - 1 \)
To find the second number, we divide the total product by the first number:
\( \text{Second number} = \frac{16x^4 - 1}{2x - 1} \)
We perform polynomial division of \( 16x^4 - 1 \) by \( 2x - 1 \) using zero placeholders for the missing intermediate powers:
\[ \begin{array}{rll} 8x^3 + 4x^2 + 2x + 1 && \\ 2x - 1 \ \overline{\smash{)} \ 16x^4 + 0x^3 + 0x^2 + 0x - 1} && \\ \underline{16x^4 - 8x^3} && \\ 8x^3 + 0x^2 + 0x - 1 && \\ \underline{8x^3 - 4x^2} && \\ 4x^2 + 0x - 1 && \\ \underline{4x^2 - 2x} && \\ 2x - 1 && \\ \underline{2x - 1} && \\ 0 && \end{array} \] Thus, the other number is \( 8x^3 + 4x^2 + 2x + 1 \).
In simple words: If you know the product of two numbers and one of them, divide the product by that number to find the other.

Exam Tip: Alternatively, you can use the algebraic identity \( A^2 - B^2 = (A - B)(A + B) \) to factorise \( 16x^4 - 1 \) as \( (4x^2 - 1)(4x^2 + 1) \), and factorise further to easily check your answer.

Question 5. Divide \( x^6 - y^6 \) by the product of \( x^2 + xy + y^2 \) and \( x - y \).
Answer:
First, we find the product of the two divisor expressions:
\( \text{Product} = (x - y)(x^2 + xy + y^2) \)
Multiplying term by term:
\( = x(x^2 + xy + y^2) - y(x^2 + xy + y^2) \)
\( = x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3 \)
Combining like terms:
\( = x^3 - y^3 \)

Now, we divide \( x^6 - y^6 \) by this product \( x^3 - y^3 \):
\( \frac{x^6 - y^6}{x^3 - y^3} \)
Using polynomial long division of \( x^6 - y^6 \) by \( x^3 - y^3 \): \[ \begin{array}{rll} x^3 + y^3 && \\ x^3 - y^3 \ \overline{\smash{)} \ x^6 - y^6} && \\ \underline{x^6 - x^3y^3} && \\ x^3y^3 - y^6 && \\ \underline{x^3y^3 - y^6} && \\ 0 && \end{array} \] Thus, the result of the division is \( x^3 + y^3 \).
In simple words: First, multiply the two smaller expressions together to get \( x^3 - y^3 \). Then, divide the larger expression \( x^6 - y^6 \) by this result.

Exam Tip: Notice that \( x^6 - y^6 \) is a difference of squares: \( (x^3)^2 - (y^3)^2 \). Factoring it directly can save you from having to do long division.

Simplification

Simplifying Algebraic Expressions by Removing Brackets

To simplify expressions containing multiple brackets, we must resolve them in a specific order. The various brackets used in algebra are listed below from first to last to be solved:

1. Vinculum or bar bracket: \( \overline{\quad} \)
2. Parentheses or small brackets: \( ( \ ) \)
3. Braces or curly brackets: \( \{ \ \} \)
4. Square or big brackets: \( [ \ ] \)

When working on a complex expression, always remove the brackets in this exact sequence: first vinculum, then small brackets, then curly brackets, and finally square brackets.

Exercise 11(E)

Question 1. Simplify : \( a^2 - 2a + \{5a^2 - (3a - 4a^2)\} \)
Answer:
First, we remove the inner parentheses \( ( \ ) \):
\( = a^2 - 2a + \{5a^2 - 3a + 4a^2\} \)
Now, combine the like terms inside the curly braces \( \{ \ \} \):
\( = a^2 - 2a + \{9a^2 - 3a\} \)
Next, we remove the curly braces:
\( = a^2 - 2a + 9a^2 - 3a \)
Finally, group and add the like terms together:
\( = (a^2 + 9a^2) + (-2a - 3a) \)
\( = 10a^2 - 5a \)
In simple words: First open the small inner brackets. Then add the terms that are alike inside the curly brackets. Lastly, open the curly brackets and combine everything.

Exam Tip: Remember that a minus sign in front of a bracket changes the sign of every term inside it when you open the bracket.

Question 2. Simplify : \( x - y - \{x - y - (x + y) - \overline{x - y}\} \)
Answer:
First, remove the vinculum (bar bracket):
\( = x - y - \{x - y - (x + y) - (x - y)\} \)
Next, open the inner parentheses:
\( = x - y - \{x - y - x - y - x + y\} \)
Simplify the expression inside the curly braces by combining like terms:
\( = x - y - \{-x - y\} \)
Now, remove the curly braces (be careful with the negative sign outside):
\( = x - y + x + y \)
Combine the remaining terms:
\( = 2x \)
In simple words: Solve the bar bracket first, which acts like a small parenthesis. Then open the parentheses, simplify inside the curly brackets, and finish by adding the remaining like terms.

Exam Tip: Treat the vinculum exactly like a set of parentheses. Any sign directly in front of the bar applies to the entire expression under the bar.

Question 3. Simplify : \( -3(1 - x^2) - 2\{x^2 - (3 - 2x^2)\} \)
Answer:
First, we expand the first term and open the inner parentheses inside the braces:
\( = -3 + 3x^2 - 2\{x^2 - 3 + 2x^2\} \)
Now, simplify the terms inside the curly braces:
\( = -3 + 3x^2 - 2\{3x^2 - 3\} \)
Multiply the terms inside the braces by \( -2 \):
\( = -3 + 3x^2 - 6x^2 + 6 \)
Finally, combine the constant terms and the \( x^2 \) terms:
\( = (-3 + 6) + (3x^2 - 6x^2) \)
\( = 3 - 3x^2 \)
In simple words: First multiply \( -3 \) into the first bracket, and open the small bracket inside the curly braces. Then simplify inside the braces before multiplying them by \( -2 \). Finally, group the numbers and the \( x^2 \) parts together.

Exam Tip: Do not forget to multiply the entire parenthesis or brace by the number in front of it, including its sign.

Question 4. Simplify : \( 2\{m - 3(n + \overline{m - 2n})\} \)
Answer:
First, remove the vinculum (bar bracket):
\( = 2\{m - 3(n + m - 2n)\} \)
Combine the like terms inside the parentheses:
\( = 2\{m - 3(m - n)\} \)
Now, multiply the parentheses by \( -3 \):
\( = 2\{m - 3m + 3n\} \)
Simplify inside the curly braces:
\( = 2\{-2m + 3n\} \)
Finally, expand the expression by multiplying by \( 2 \):
\( = -4m + 6n \)
\( = 6n - 4m \)
In simple words: Start by removing the bar over the last terms. Combine the \( n \) terms inside the parentheses, multiply them by \( -3 \), simplify inside the curly brackets, and double everything at the end.

Exam Tip: Keep your final algebraic expressions ordered alphabetically (e.g., write \( -4m + 6n \) or rearrange as \( 6n - 4m \)) for standard presentation.

Question 5. Simplify : \( 3x - [3x - \{3x - (3x - \overline{3x - y})\}] \)
Answer:
First, remove the bar bracket (vinculum):
\( = 3x - [3x - \{3x - (3x - 3x + y)\}] \)
Simplify inside the parentheses:
\( = 3x - [3x - \{3x - (y)\}] \)
Now, remove the curly braces:
\( = 3x - [3x - \{3x - y\}] \)
\( = 3x - [3x - 3x + y] \)
Simplify inside the square brackets:
\( = 3x - [y] \)
Finally, remove the square brackets:
\( = 3x - y \)
In simple words: Work your way from the inside out: first remove the bar, then simplify the parentheses, then the curly brackets, and finally the square brackets.

Exam Tip: In nested bracket problems, work methodically from the innermost bracket outward. Jumping steps can easily lead to a sign error.

Question 6. Simplify : \( p^2x - 2\{px - 3x(x^2 - \overline{3a - x^2})\} \)
Answer:
First, remove the bar bracket:
\( = p^2x - 2\{px - 3x(x^2 - 3a + x^2)\} \)
Simplify the terms inside the parentheses:
\( = p^2x - 2\{px - 3x(2x^2 - 3a)\} \)
Now, expand the parentheses by multiplying by \( -3x \):
\( = p^2x - 2\{px - 6x^3 + 9ax\} \)
Finally, expand the curly braces by multiplying every term inside by \( -2 \):
\( = p^2x - 2px + 12x^3 - 18ax \)
In simple words: Start by clearing the bar bracket. Combine the \( x^2 \) terms inside the parenthesis, then multiply that bracket by \( -3x \). Lastly, multiply all terms inside the curly braces by \( -2 \).

Exam Tip: Be very careful when multiplying \( -3x \) by \( -3a \) - the product is positive (\( +9ax \)). Always verify sign changes when distributing a negative term.

Question 7. Simplify : \( 2[6 + 4\{m - 6(7 - \overline{n + p}) + q\}] \)
Answer:
First, remove the vinculum (bar bracket):
\( = 2[6 + 4\{m - 6(7 - n - p) + q\}] \)
Now, expand the parentheses by multiplying by \( -6 \):
\( = 2[6 + 4\{m - 42 + 6n + 6p + q\}] \)
Next, expand the curly braces by multiplying all terms inside by \( 4 \):
\( = 2[6 + 4m - 168 + 24n + 24p + 4q] \)
Simplify the numerical terms inside the square brackets:
\( = 2[4m + 24n + 24p + 4q - 162] \)
Finally, multiply everything inside the square brackets by \( 2 \):
\( = 8m + 48n + 48p + 8q - 324 \)
In simple words: Clear the bar bracket first. Then multiply the parentheses by \( -6 \), multiply the braces by \( 4 \), group the constant numbers together, and double the entire expression at the end.

Exam Tip: Keep track of the signs during multiple distributions. For example, \( -6 \times -n = +6n \) and \( -6 \times -p = +6p \).

Question 8. Simplify : \( a - [a - \overline{b + a} - \{a - (a - \overline{b - a})\}] \)
Answer:
First, remove the vinculums (bar brackets):
\( = a - [a - b - a - \{a - (a - b + a)\}] \)
Simplify the terms inside the square brackets and parentheses:
\( = a - [-b - \{a - (2a - b)\}] \)
Remove the parentheses inside the curly braces:
\( = a - [-b - \{a - 2a + b\}] \)
\( = a - [-b - \{-a + b\}] \)
Now, remove the curly braces:
\( = a - [-b + a - b] \)
Combine the like terms inside the square brackets:
\( = a - [a - 2b] \)
Finally, remove the square brackets:
\( = a - a + 2b \)
\( = 2b \)
In simple words: Clear both of the bar brackets first. Next, simplify the terms inside the parentheses and braces, and then open them one by one until you can combine the final \( a \) and \( b \) terms.

Exam Tip: Make sure you simplify inside each bracket as much as possible before removing it. This reduces the number of terms you have to manage and prevents silly mistakes.

Question 9. Simplify : \( 3x - [4x - \overline{3x - 5y} - 3\{2x - (3x - \overline{2x - 3y})\}] \)
Answer:
First, remove the vinculums (bar brackets):
\( = 3x - [4x - 3x + 5y - 3\{2x - (3x - 2x + 3y)\}] \)
Simplify inside the parentheses and the first part of the square brackets:
\( = 3x - [x + 5y - 3\{2x - (x + 3y)\}] \)
Remove the parentheses inside the curly braces:
\( = 3x - [x + 5y - 3\{2x - x - 3y\}] \)
Simplify inside the curly braces:
\( = 3x - [x + 5y - 3\{x - 3y\}] \)
Now, expand the curly braces by multiplying by \( -3 \):
\( = 3x - [x + 5y - 3x + 9y] \)
Simplify inside the square brackets:
\( = 3x - [-2x + 14y] \)
Remove the square brackets:
\( = 3x + 2x - 14y \)
\( = 5x - 14y \)
In simple words: Get rid of the bars first. Then simplify the inside parts of the parentheses, multiply them out, combine like terms within each set of brackets, and finish by adding the final terms.

Exam Tip: Double-check the signs when removing brackets. For instance, the expression \( -3(x - 3y) \) expands to \( -3x + 9y \), not \( -3x - 9y \).

Question 10. Simplify : \( a^5 \div a^3 + 3a \times 2a \)
Answer:
Following the order of operations, we perform division and multiplication first from left to right:
For the division: \( a^5 \div a^3 = a^{5-3} = a^2 \)
For the multiplication: \( 3a \times 2a = 6a^2 \)
Now, add the two results together:
\( = a^2 + 6a^2 \)
\( = 7a^2 \)
In simple words: First divide the powers of \( a \) by subtracting their exponents. Then multiply the other two terms. Finally, add the like terms together.

Exam Tip: Remember the DMAS rule (Division, Multiplication, Addition, Subtraction). Division and multiplication must be done before doing addition.

Question 11. Simplify : \( x^5 \div (x^2 \times y^2) \times y^3 \)
Answer:
First, resolve the product inside the parentheses:
\( = x^5 \div (x^2 y^2) \times y^3 \)
Write the division as a fraction:
\( = \frac{x^5}{x^2 y^2} \times y^3 \)
Apply the laws of exponents to simplify the \( x \) and \( y \) terms:
\( = x^{5-2} \cdot y^{-2} \cdot y^3 \)
\( = x^3 y^{3-2} \)
\( = x^3 y \)
In simple words: Combine the variables inside the brackets first, then divide \( x^5 \) by that result, and multiply by \( y^3 \) using the subtraction and addition rules of exponents.

Exam Tip: When dividing variables with exponents, subtract the exponent of the denominator from that of the numerator: \( \frac{x^a}{x^b} = x^{a-b} \).

Question 12. Simplify : \( (x^5 \div x^2) \times y^2 \times y^3 \)
Answer:
First, simplify the division inside the parentheses:
\( = x^{5-2} \times y^2 \times y^3 \)
\( = x^3 \times y^2 \times y^3 \)
Now, multiply the \( y \) terms by adding their exponents:
\( = x^3 y^{2+3} \)
\( = x^3 y^5 \)
In simple words: First solve the division inside the brackets by subtracting the powers of \( x \). Then multiply the \( y \) parts by adding their powers together.

Exam Tip: When multiplying identical bases, add their exponents together (\( y^a \times y^b = y^{a+b} \)), but keep different bases (like \( x \) and \( y \)) separate.

Question 13. Simplify : \( (y^3 - 5y^2) \div y \times (y - 1) \)
Answer:
First, divide the term inside the parentheses by \( y \):
\( = \left( \frac{y^3 - 5y^2}{y} \right) \times (y - 1) \)
\( = (y^2 - 5y) \times (y - 1) \)
Next, multiply the two binomials:
\( = y^2(y - 1) - 5y(y - 1) \)
\( = y^3 - y^2 - 5y^2 + 5y \)
Finally, combine the like terms \( -y^2 \) and \( -5y^2 \):
\( = y^3 - 6y^2 + 5y \)
In simple words: First, divide both parts inside the first bracket by \( y \). Then, multiply the result by the second bracket term-by-term and group the \( y^2 \) terms together.

Exam Tip: When dividing a polynomial by a monomial, ensure you divide every single term of the polynomial by the monomial.

Question 14. Simplify : \( 3a \times [8b \div 4 - 6\{a - (5a - \overline{3b - 2a})\}] \)
Answer:
First, remove the vinculum (bar bracket):
\( = 3a \times [8b \div 4 - 6\{a - (5a - 3b + 2a)\}] \)
Perform the division at the beginning of the square brackets and simplify terms inside the parentheses:
\( = 3a \times [2b - 6\{a - (7a - 3b)\}] \)
Remove the parentheses:
\( = 3a \times [2b - 6\{a - 7a + 3b\}] \)
Simplify inside the curly braces:
\( = 3a \times [2b - 6\{-6a + 3b\}] \)
Multiply the terms inside the braces by \( -6 \):
\( = 3a \times [2b + 36a - 18b] \)
Simplify inside the square brackets:
\( = 3a \times [36a - 16b] \)
Finally, expand the expression by multiplying by \( 3a \):
\( = 108a^2 - 48ab \)
In simple words: Solve the bar over \( 3b - 2a \) first, then simplify inside the parentheses and curly braces. Multiply the remaining parts inside the square brackets by \( -6 \), combine \( b \) terms, and multiply by \( 3a \) at the very end.

Exam Tip: Solve operations inside different parts of the expression simultaneously if they are independent, like \( 8b \div 4 \), to save time.

Question 15. Simplify : \( 7x + 4\{x^2 \div (5x \div 10)\} - 3\{2 - x^3 \div (3x^2 \div x)\} \)
Answer:
First, simplify the division operations inside the innermost parentheses:
For the first parenthesis: \( 5x \div 10 = \frac{5x}{10} = \frac{x}{2} \)
For the second parenthesis: \( 3x^2 \div x = 3x \)
Substitute these back into the expression:
\( = 7x + 4\left\{x^2 \div \frac{x}{2}\right\} - 3\left\{2 - x^3 \div 3x\right\} \)
Now, simplify the divisions inside the curly braces:
\( x^2 \div \frac{x}{2} = x^2 \times \frac{2}{x} = 2x \)
\( x^3 \div 3x = \frac{x^3}{3x} = \frac{x^2}{3} \)
Substitute these back:
\( = 7x + 4\{2x\} - 3\left\{2 - \frac{x^2}{3}\right\} \)
Now, expand and remove the curly braces:
\( = 7x + 8x - 6 + x^2 \)
Combine like terms and rearrange in descending order:
\( = x^2 + 15x - 6 \)
In simple words: First simplify the fractions inside the small parentheses. Then, perform the divisions inside the curly braces. Finally, multiply the terms out, combine the \( x \) terms, and write the answer in order of powers.

Exam Tip: When dividing by a fraction, multiply by its reciprocal: \( x^2 \div \frac{x}{2} = x^2 \times \frac{2}{x} \). Be extremely careful with signs when distributing negative coefficients.