Selina Concise Solutions for ICSE Class 6 Mathematics Chapter 20 Substitution

Substitution - Important Points

1. Substitution: The total value of an algebraic expression changes depending on what number we put in place of its variables. 2. Using Brackets: We use symbols like line bars, parentheses, curly brackets, and square brackets, which are written as \( \overline{\quad} \), \( ( ) \), \( \{ \} \), and \( [ ] \). When a group of terms is put inside a bracket, we treat it as a single group. Points to remember: - When we make an expression simpler, we must solve the terms inside the brackets first. - We call \( ( ) \) small brackets or parentheses. - We call \( \{ \} \) middle brackets or curly brackets. - We call \( [ ] \) big brackets or square brackets. - If we need another bracket, we draw a line over the terms. This line is a bar bracket, also known as a vinculum. For example, in \( 3x + \overline{4y - 5z} \), the line above \( 4y - 5z \) is the bar bracket.

Exercise 20(A)

Question 1. Fill in the following blanks, when :
\( x = 3, y = 6, z = 18, a = 2, b = 8, c = 32 \) and \( d = 0 \).
(i) \( x + y = \) ................
(ii) \( y - x = \) ................
(iii) \( \frac{y}{x} = \) ................
(iv) \( c \div b = \) ................
(v) \( z \div x = \) ................
(vi) \( y \times d = \) ................
(vii) \( d \div x = \) ................
(viii) \( ab + y = \) ................
(ix) \( a + b + x = \) ................
(x) \( b + z - d = \) ................
(xi) \( a - b + y = \) ................
(xii) \( z - a - b = \) ................
(xiii) \( d - a + x = \) ................
(xiv) \( xy - bd = \) ................
(xv) \( xz + cd = \) ................
Answer:
(i) \( x + y = 3 + 6 = 9 \)
(ii) \( y - x = 6 - 3 = 3 \)
(iii) \( \frac{y}{x} = \frac{6}{3} = 2 \)
(iv) \( c \div b = \frac{c}{b} = \frac{32}{8} = 4 \)
(v) \( z \div x = \frac{z}{x} = \frac{18}{3} = 6 \)
(vi) \( y \times d = 6 \times 0 = 0 \)
(vii) \( d \div x = \frac{d}{x} = \frac{0}{3} = 0 \)
(viii) \( ab + y = 2 \times 8 + 6 = 16 + 6 = 22 \)
(ix) \( a + b + x = 2 + 8 + 3 = 13 \)
(x) \( b + z - d = 8 + 18 - 0 = 26 \)
(xi) \( a - b + y = 2 - 8 + 6 = 8 - 8 = 0 \)
(xii) \( z - a - b = 18 - 2 - 8 = 18 - 10 = 8 \)
(xiii) \( d - a + x = 0 - 2 + 3 = 1 \)
(xiv) \( xy - bd = 3 \times 6 - 8 \times 0 = 18 - 0 = 18 \)
(xv) \( xz + cd = 3 \times 18 + 32 \times 0 = 54 + 0 = 54 \)
In simple words: To find the answer, just put the given numbers in place of the letters and calculate.

Exam Tip: Be very careful when putting zero into these equations, since multiplying by zero always gives zero.

Question 2. Find the value of :
(i) \( p + 2q + 3r \), when \( p = 1, q = 5 \) and \( r = 2 \)
(ii) \( 2a + 4b + 5c \), when \( a = 5, b = 10 \) and \( c = 20 \)
(iii) \( 3a - 2b \), when \( a = 8 \) and \( b = 10 \)
(iv) \( 5x + 3y - 6z \), when \( x = 3, y = 5 \) and \( z = 4 \)
(v) \( 2p - 3q + 4r - 8s \), when \( p = 10, q = 8, r = 6, \) and \( s = 2 \)
(vi) \( 6m - 2n - 5p - 3q \), when \( m = 20, n = 10, p = 2 \) and \( q = 9 \)
Answer:
(i) \( p + 2q + 3r = 1 + 2(5) + 3(2) = 1 + 10 + 6 = 17 \)
(ii) \( 2a + 4b + 5c = 2(5) + 4(10) + 5(20) = 10 + 40 + 100 = 150 \)
(iii) \( 3a - 2b = 3(8) - 2(10) = 24 - 20 = 4 \)
(iv) \( 5x + 3y - 6z = 5(3) + 3(5) - 6(4) = 15 + 15 - 24 = 30 - 24 = 6 \)
(v) \( 2p - 3q + 4r - 8s = 2(10) - 3(8) + 4(6) - 8(2) = 20 - 24 + 24 - 16 = 20 - 16 = 4 \)
(vi) \( 6m - 2n - 5p - 3q = 6(20) - 2(10) - 5(2) - 3(9) = 120 - 20 - 10 - 27 = 120 - 57 = 63 \)
In simple words: Change the letters into the given numbers and multiply first, then add and subtract.

Exam Tip: Remember to use BODMAS rules, doing multiplication before addition or subtraction.

Question 3. Find the value of :
(i) \( 4pq \times 2r \), when \( p = 5, q = 3 \) and \( r = 1/2 \)
(ii) \( \frac{yx}{z} \), when \( x = 8, y = 4 \) and \( z = 16 \)
(iii) \( \frac{a + b - c}{2a} \), when \( a = 5, b = 7 \) and \( c = 2 \)
Answer:
(i) \( 4pq \times 2r = 4 \times 5 \times 3 \times 2 \times \frac{1}{2} = 60 \)
(ii) \( \frac{yx}{z} = \frac{4 \times 8}{16} = \frac{32}{16} = 2 \)
(iii) \( \frac{a + b - c}{2a} = \frac{5 + 7 - 2}{2 \times 5} = \frac{12 - 2}{10} = \frac{10}{10} = 1 \)
In simple words: Replace the variables with their fractions or numbers, then simplify the fractions step by step.

Exam Tip: Simplify expressions with fractions carefully, especially when canceling out values in both the numerator and denominator.

Question 4. If \( a = 3, b = 0, c = 2 \) and \( d = 1 \), find the value of :
(i) \( 3a + 2b - 6c + 4d \)
(ii) \( 6a - 3b - 4c - 2d \)
(iii) \( ab - bc + cd - da \)
(iv) \( abc - bcd + cda \)
(v) \( a^2 + 2b^2 - 3c^2 \)
(vi) \( a^2 + b^2 - c^2 + d^2 \)
(vii) \( 2a^2 - 3b^2 + 4c^2 - 5d^2 \)
Answer:
(i) \( 3(3) + 2(0) - 6(2) + 4(1) = 9 + 0 - 12 + 4 = 1 \)
(ii) \( 6(3) - 3(0) - 4(2) - 2(1) = 18 - 0 - 8 - 2 = 8 \)
(iii) \( (3)(0) - (0)(2) + (2)(1) - (1)(3) = 0 - 0 + 2 - 3 = -1 \)
(iv) \( (3)(0)(2) - (0)(2)(1) + (2)(1)(3) = 0 - 0 + 6 = 6 \)
(v) \( 3^2 + 2(0^2) - 3(2^2) = 9 + 0 - 3(4) = 9 - 12 = -3 \)
(vi) \( 3^2 + 0^2 - 2^2 + 1^2 = 9 + 0 - 4 + 1 = 6 \)
(vii) \( 2(3^2) - 3(0^2) + 4(2^2) - 5(1^2) = 2(9) - 0 + 4(4) - 5(1) = 18 + 16 - 5 = 29 \)
In simple words: First square the numbers if there is a power of two, then multiply and solve the remaining parts.

Exam Tip: Remember that any variable with a value of zero will make that whole term zero when multiplied.

Question 5. Find the value of \( 5x^2 - 3x + 2 \), when \( x = 2 \).
Answer:
\( 5x^2 - 3x + 2 = 5(2)^2 - 3(2) + 2 \)
\( = 5(4) - 6 + 2 \)
\( = 20 - 6 + 2 = 16 \)
In simple words: Swap \( x \) with \( 2 \) in the expression, square \( 2 \) first, then solve the equation.

Exam Tip: Always calculate the power before doing any multiplication with the coefficient.

Question 6. Find the value of \( 3x^3 - 4x^2 + 5x - 6 \), when \( x = -1 \).
Answer:
\( 3x^3 - 4x^2 + 5x - 6 = 3(-1)^3 - 4(-1)^2 + 5(-1) - 6 \)
\( = 3(-1) - 4(1) - 5 - 6 \)
\( = -3 - 4 - 5 - 6 = -18 \)
In simple words: Swap \( x \) with \( -1 \). Remember that an odd power keeps the negative sign, while an even power turns it positive.

Exam Tip: Pay extra attention to negative bases raised to odd or even powers, as this is a common place to make sign mistakes.

Question 7. Show that the value of \( x^3 - 8x^2 + 12x - 5 \) is zero, when \( x = 1 \).
Answer:
Substituting \( x = 1 \):
\( x^3 - 8x^2 + 12x - 5 = (1)^3 - 8(1)^2 + 12(1) - 5 \)
\( = 1 - 8 + 12 - 5 \)
\( = 13 - 13 = 0 \)
This confirms that the result is indeed zero.
In simple words: Put \( 1 \) in place of \( x \) and combine the positive and negative numbers to get zero.

Exam Tip: Grouping positive and negative numbers separately before adding them makes calculations easier and prevents arithmetic errors.

Question 8. State true and false :
(i) The value of \( x + 5 = 6 \), when \( x = 1 \)
(ii) The value of \( 2x - 3 = 1 \), when \( x = 0 \)
(iii) \( \frac{2x - 4}{x + 1} = -1 \), when \( x = 1 \)
Answer:
(i) True. Putting \( x = 1 \) gives:
\( 1 + 5 = 6 \)
\( 6 = 6 \) (Verified)
(ii) False. Putting \( x = 0 \) gives:
\( 2(0) - 3 = -3 \), which does not equal \( 1 \).
(iii) True. Putting \( x = 1 \) gives:
\( \frac{2(1) - 4}{1 + 1} = \frac{2 - 4}{2} = \frac{-2}{2} = -1 \) (Verified)
In simple words: Put the given value of \( x \) into each equation to check if the left side equals the right side.

Exam Tip: Always write down the calculations clearly for verification questions instead of just writing True or False.

Question 9. If \( x = 2, y = 5 \) and \( z = 4 \), find the value of each of the following :
(i) \( \frac{x}{2x^2} \)
(ii) \( \frac{xz}{yz} \)
(iii) \( z^x \)
(iv) \( y^x \)
(v) \( \frac{x^2y^2z^2}{xz} \)
(vi) \( \frac{5x^4y^2z^2}{2x^2} \)
(vii) \( \frac{xy}{y^2z} \)
(viii) \( \frac{x^2y^x}{x} \)
Answer:
(i) \( \frac{x}{2x^2} = \frac{2}{2(2)^2} = \frac{2}{2 \times 4} = \frac{1}{4} \)
(ii) \( \frac{xz}{yz} = \frac{2 \times 4}{5 \times 4} = \frac{2}{5} \)
(iii) \( z^x = 4^2 = 4 \times 4 = 16 \)
(iv) \( y^x = 5^2 = 5 \times 5 = 25 \)
(v) \( \frac{x^2y^2z^2}{xz} = \frac{(2)^2 \times (5)^2 \times (4)^2}{2 \times 4} = 2^{2-1} \times 5^2 \times 4^{2-1} = 2 \times 25 \times 4 = 200 \)
(vi) \( \frac{5x^4y^2z^2}{2x^2} = \frac{5x^{4-2}y^2z^2}{2} = \frac{5x^2y^2z^2}{2} = \frac{5(2)^2(5)^2(4)^2}{2} = \frac{5 \times 4 \times 25 \times 16}{2} = 4000 \)
(vii) \( \frac{xy}{y^2z} = \frac{x}{y^{2-1}z} = \frac{x}{yz} = \frac{2}{5 \times 4} = \frac{1}{10} \)
(viii) \( \frac{x^2y^x}{x} = x^{2-1}y^x = xy^x = (2)(5)^2 = 2 \times 25 = 50 \)
In simple words: Simplify the powers of letters before putting in the numbers to make the work much easier.

Exam Tip: Applying the rules of exponents to simplify the expression first will save you from multiplying large numbers.

Question 10. If \( a = 3, find the values of \( a^2 \) and \( 2^a \).
Answer:
\( a^2 = (3)^2 = 3 \times 3 = 9 \)
\( 2^a = (2)^3 = 2 \times 2 \times 2 = 8 \)
In simple words: \( a^2 \) means multiplying \( 3 \) by itself. \( 2^a \) means multiplying \( 2 \) three times.

Exam Tip: Do not confuse the base and the exponent, as \( a^2 \) and \( 2^a \) mean very different things.

Question 11. If \( m = 2 \), find the difference between the values of \( 4m^3 \) and \( 3m^4 \).
Answer:
First, find the value of each term:
\( 4m^3 = 4(2)^3 = 4 \times 8 = 32 \)
\( 3m^4 = 3(2)^4 = 3 \times 16 = 48 \)
Now, find the difference between these two values:
\( 48 - 32 = 16 \)
In simple words: Find both values by substituting \( m = 2 \), then subtract the smaller number from the larger one.

Exam Tip: "Difference" always refers to the absolute subtraction of the smaller value from the larger value to get a positive result.

Exercise 20(B)

Question 1. Evaluate :
(i) \( (23 - 15) + 4 \)
(ii) \( 5x + (3x + 7x) \)
(iii) \( 6m - (4m - m) \)
(iv) \( (9a - 3a) + 4a \)
(v) \( 35b - (16b + 9b) \)
(vi) \( (3y + 8y) - 5y \)
Answer:
(i) \( (23 - 15) + 4 = 8 + 4 = 12 \)
(ii) \( 5x + (3x + 7x) = 5x + 10x = 15x \)
(iii) \( 6m - (4m - m) = 6m - 3m = 3m \)
(iv) \( (9a - 3a) + 4a = 6a + 4a = 10a \)
(v) \( 35b - (16b + 9b) = 35b - 25b = 10b \)
(vi) \( (3y + 8y) - 5y = 11y - 5y = 6y \)
In simple words: Work out the numbers inside the brackets first, then add or subtract the rest.

Exam Tip: Solve the expression inside the brackets first to follow the correct mathematical order of operations.

Question 2. Simplify :
(i) \( 12x - (5x + 2x) \)
(ii) \( 10m + (4n - 3n) - 5n \)
(iii) \( (15b - 6b) - (8b + 4b) \)
(iv) \( -(-4a - 8a) \)
(v) \( x - (x - y) - (-x + y) \)
(vi) \( p + (-q - r - s) - (p - q - r) \)
(vii) \( (a + b) - (c + d) - (e - f) \)
(viii) \( 3x + (8x - 5x) - (7x - x) \)
(ix) \( a - (a - b - c) \)
(x) \( 6a^2 + (2a^2 - a^2) - (a^2 - b^2) \)
(xi) \( 2m - (3m + 2n - 6n) \)
(xii) \( -m - n - (-m) - m \)
(xiii) \( x + y - \overline{x + y - x} \)
(xiv) \( 25y - (5x - 10y + 6x - 3y) \)
(xv) \( 3x + \overline{2x - x} + 2 \)
(xvi) \( a - \overline{2a - 4a} + 3a \)
(xvii) \( 5x^2 - (3x - \overline{x^2 - 4}) \)
(xviii) \( -(y - x) - (x + y - \overline{2x + y}) \)
Answer:
(i) \( 12x - 7x = 5x \)
(ii) \( 10m + n - 5n = 10m - 4n \)
(iii) \( 9b - 12b = -3b \)
(iv) \( -(-12a) = 12a \)
(v) \( x - x + y + x - y = x \)
(vi) \( p - q - r - s - p + q + r = -s \)
(vii) \( a + b - c - d - e + f \)
(viii) \( 3x + 3x - 6x = 6x - 6x = 0 \)
(ix) \( a - a + b + c = b + c \)
(x) \( 6a^2 + a^2 - a^2 + b^2 = 6a^2 + b^2 \)
(xi) \( 2m - (3m - 4n) = 2m - 3m + 4n = -m + 4n \)
(xii) \( -m - n + m - m = -m - n \)
(xiii) \( x + y - (x + y - x) = x + y - y = x \)
(xiv) \( 25y - (11x - 13y) = 25y - 11x + 13y = 38y - 11x \)
(xv) \( 3x + (2x - x) - 2 = 3x + x - 2 = 4x - 2 \)
(xvi) \( a - (2a - 4a) - 3a = a - 2a + 4a + 3a = 8a - 2a = 6a \)
(xvii) \( 5x^2 - (3x - x^2 + 4) = 5x^2 - 3x + x^2 - 4 = 6x^2 - 3x - 4 \)
(xviii) \( -y + x - (x + y - 2x - y) = -y + x - (-x) = -y + x + x = 2x - y \)
In simple words: When there is a minus sign outside a bracket or bar, it reverses the signs of all terms inside when you remove it.

Exam Tip: A bar bracket (vinculum) acts like a parenthesis. When removing brackets with a negative sign outside, always change the signs inside.

Question 3. Simplify :
(i) \( x - (y - z) + x + (y - z) + y - (z + x) \)
(ii) \( x - [y + \{x - (y + x)\}] \)
(iii) \( 4x + 3(2x - 5y) \)
(iv) \( 2(3a - b) - 5(a - 3b) \)
(v) \( p + 2(q - \overline{r + p}) \)
(vi) \( a - [ -\{-(a - \overline{b - c})\} ] \)
(vii) \( 3x - [5y - \{6y + 2(10y - x)\}] \)
(viii) \( 5\{a^2 - a(a - \overline{a - 2})\} \)
Answer:
(i) \( x - y + z + x + y - z + y - z - x = x + y - z \)
(ii) \( x - [y + \{x - y - x\}] = x - [y - y] = x \)
(iii) \( 4x + 6x - 15y = 10x - 15y \)
(iv) \( 6a - 2b - 5a + 15b = a + 13b \)
(v) \( p + 2(q - r - p) = p + 2q - 2r - 2p = 2q - 2r - p \)
(vi) \( a - [ -\{-(a - b + c)\} ] = a - [ -\{-a + b - c\} ] = a - [a - b + c] = b - c \)
(vii) \( 3x - [5y - \{26y - 2x\}] = 3x - [5y - 26y + 2x] = 3x - [-21y + 2x] = x + 21y \)
(viii) \( 5\{a^2 - a(a - a + 2)\} = 5\{a^2 - 2a\} = 5a^2 - 10a \)
In simple words: Solve from the innermost brackets first, starting with the bar brackets, then parentheses, curly brackets, and finally square brackets.

Exam Tip: Work step-by-step from inside to outside when simplifying nested brackets to keep from losing track of negative signs.

Exercise 20(C)

Question 1. Fill in the blanks :
(i) \( 2a + b - c = 2a + (\text{..........}) \)
(ii) \( 3x - z + y = 3x - (\text{..........}) \)
(iii) \( 6p - 5x + q = 6p - (\text{..........}) \)
(iv) \( a + b - c + d = a + (\text{..........}) \)
(v) \( 5a + 4b + 4x - 2c = 4x - (\text{..........}) \)
(vi) \( 7x + 2z + 4y - 3 = -3 + 4y + (\text{..........}) \)
(vii) \( 3m - 2n + 6 = 6 - (\text{..........}) \)
(viii) \( 2t + r - p - q + s = 2t + r - (\text{..........}) \)
Answer:
(i) \( 2a + b - c = 2a + (b - c) \)
(ii) \( 3x - z + y = 3x - (z - y) \)
(iii) \( 6p - 5x + q = 6p - (5x - q) \)
(iv) \( a + b - c + d = a + (b - c + d) \)
(v) \( 5a + 4b + 4x - 2c = 4x - (2c - 5a - 4b) \)
(vi) \( 7x + 2z + 4y - 3 = -3 + 4y + (7x + 2z) \)
(vii) \( 3m - 2n + 6 = 6 - (2n - 3m) \)
(viii) \( 2t + r - p - q + s = 2t + r - (p + q - s) \)
In simple words: When putting terms into a bracket with a plus sign outside, keep their signs the same. If there is a minus sign outside, switch all their signs.

Exam Tip: Always double check your signs by expanding the bracket to see if you get the original expression back.

Question 2. Insert the bracket as indicated :
(i) \( x - 2y = - (\text{...............}) \)
(ii) \( m + n - p = - (\text{...............}) \)
(iii) \( a + 4b - 4c = a + (\text{...............}) \)
(iv) \( a - 3b + 5c = a - (\text{...............}) \)
(v) \( x^2 - y^2 + z^2 = x^2 - (\text{...............}) \)
(vi) \( m^2 + x^2 - p^2 = - (\text{...............}) \)
(vii) \( 2x - y + 2z = 2z - (\text{...............}) \)
(viii) \( ab + 2bc - 3ac = 2bc - (\text{...............}) \)
Answer:
(i) \( x - 2y = -(2y - x) \)
(ii) \( m + n - p = -(p - m - n) \)
(iii) \( a + 4b - 4c = a + (4b - 4c) \)
(iv) \( a - 3b + 5c = a - (3b - 5c) \)
(v) \( x^2 - y^2 + z^2 = x^2 - (y^2 - z^2) \)
(vi) \( m^2 + x^2 - p^2 = -(p^2 - m^2 - x^2) \)
(vii) \( 2x - y + 2z = 2z - (y - 2x) \)
(viii) \( ab + 2bc - 3ac = 2bc - (3ac - ab) \)
In simple words: Re-grouping with a leading negative sign means you must switch the sign of every item inside the bracket.

Exam Tip: Be very careful when factoring out negative signs, especially with squared terms or multi-letter variables.

Revision Exercise

Question 1. Find the value of \( 3ab + 10bc - 2abc \) when \( a = 2, b = 5 \) and \( c = 8 \).
Answer:
Substituting the values of \( a, b, \) and \( c \):
\( 3ab + 10bc - 2abc = 3(2)(5) + 10(5)(8) - 2(2)(5)(8) \)
\( = 30 + 400 - 160 \)
\( = 430 - 160 = 270 \)
In simple words: Place the given numbers into the expression, multiply the groups first, then add and subtract.

Exam Tip: Clearly show each multiplication step separately to make sure you do not make a mental math error.

Question 2. If \( x = 2, y = 3 \) and \( z = 4 \), find the value of \( 3x^2 - 4y^2 + 2z^2 \).
Answer:
Substituting the given values:
\( 3x^2 - 4y^2 + 2z^2 = 3(2)^2 - 4(3)^2 + 2(4)^2 \)
\( = 3(4) - 4(9) + 2(16) \)
\( = 12 - 36 + 32 \)
\( = 44 - 36 = 8 \)
In simple words: Square each number first, then multiply by the coefficients and solve the addition and subtraction.

Exam Tip: Remember to calculate the square of each variable before multiplying by its coefficient outside the parentheses.

Question 3. If \( x = 3, y = 2 \) and \( z = 1 \); find the value of:
(i) \( x^y \)
(ii) \( y^x \)
(iii) \( 3x^2 - 5y^2 \)
(iv) \( 2x - 3y + 4z + 5 \)
(v) \( y^2 - x^2 + 6z^2 \)
(vi) \( xy + y^2z - 4zx \)
Answer:
(i) \( x^y = 3^2 = 9 \)
(ii) \( y^x = 2^3 = 8 \)
(iii) \( 3x^2 - 5y^2 = 3(3)^2 - 5(2)^2 = 3(9) - 5(4) = 27 - 20 = 7 \)
(iv) \( 2x - 3y + 4z + 5 = 2(3) - 3(2) + 4(1) + 5 = 6 - 6 + 4 + 5 = 9 \)
(v) \( y^2 - x^2 + 6z^2 = (2)^2 - (3)^2 + 6(1)^2 = 4 - 9 + 6 = 1 \)
(vi) \( xy + y^2z - 4zx = (3)(2) + (2)^2(1) - 4(1)(3) = 6 + 4 - 12 = -2 \)
In simple words: Replace each letter with its number, find the powers first, and solve step by step.

Exam Tip: Be careful with signs when subtracting a larger number from a smaller number, which results in a negative value.

Question 4. If \( P = -12x^2 - 10xy + 5y^2 \), \( Q = 7x^2 + 6xy + 2y^2 \), and \( R = 5x^2 + 2xy + 4y^2 \); find :
(i) \( P - Q \)
(ii) \( Q + P \)
(iii) \( P - Q + R \)
(iv) \( P + Q + R \)
Answer:
(i) \( P - Q = (-12x^2 - 10xy + 5y^2) - (7x^2 + 6xy + 2y^2) \)
\( = -12x^2 - 10xy + 5y^2 - 7x^2 - 6xy - 2y^2 \)
\( = -19x^2 - 16xy + 3y^2 \)
(ii) \( Q + P = (7x^2 + 6xy + 2y^2) + (-12x^2 - 10xy + 5y^2) \)
\( = 7x^2 + 6xy + 2y^2 - 12x^2 - 10xy + 5y^2 \)
\( = -5x^2 - 4xy + 7y^2 \)
(iii) \( P - Q + R = (-12x^2 - 10xy + 5y^2) - (7x^2 + 6xy + 2y^2) + (5x^2 + 2xy + 4y^2) \)
\( = -12x^2 - 10xy + 5y^2 - 7x^2 - 6xy - 2y^2 + 5x^2 + 2xy + 4y^2 \)
\( = -14x^2 - 14xy + 7y^2 \)
(iv) \( P + Q + R = (-12x^2 - 10xy + 5y^2) + (7x^2 + 6xy + 2y^2) + (5x^2 + 2xy + 4y^2) \)
\( = -12x^2 + 7x^2 + 5x^2 - 10xy + 6xy + 2xy + 5y^2 + 2y^2 + 4y^2 \)
\( = -2xy + 11y^2 \)
In simple words: Group like terms (all \( x^2 \) terms, \( xy \) terms, and \( y^2 \) terms) together, then simplify them.

Exam Tip: Always put brackets around expressions when subtracting, to ensure you distribute the negative sign to all terms inside.

Question 5. If \( x = a^2 - bc \), \( y = b^2 - ca \) and \( z = c^2 - ab \); find the value of :
(i) \( ax + by + cz \)
(ii) \( ay - bx + cz \)
Answer:
(i) \( ax + by + cz = a(a^2 - bc) + b(b^2 - ca) + c(c^2 - ab) \)
\( = a^3 - abc + b^3 - abc + c^3 - abc \)
\( = a^3 + b^3 + c^3 - 3abc \)
(ii) \( ay - bx + cz = a(b^2 - ca) - b(a^2 - bc) + c(c^2 - ab) \)
\( = ab^2 - ca^2 - a^2b + b^2c + c^3 - abc \)
In simple words: Replace \( x, y, z \) with their expressions, expand each term, and then group similar terms together.

Exam Tip: Be careful with signs when distributing negative variables, such as multiplying a term by \( -b \).

Question 6. Multiply and then evaluate :
(i) \( (4x + y) \) and \( (x - 2y) \); when \( x = 2 \) and \( y = 1 \).
(ii) \( (x^2 - y) \) and \( (xy - y^2) \); when \( x = 1 \) and \( y = 2 \).
(iii) \( (x - 2y + z) \) and \( (x - 3z) \); when \( x = -2, y = -1 \) and \( z = 1 \).
Answer:
(i) Multiplication:
\( (4x + y)(x - 2y) = 4x(x - 2y) + y(x - 2y) \)
\( = 4x^2 - 8xy + xy - 2y^2 = 4x^2 - 7xy - 2y^2 \)
Verification:
L.H.S. \( = (4(2) + 1)(2 - 2(1)) = (9)(0) = 0 \)
R.H.S. \( = 4(2)^2 - 7(2)(1) - 2(1)^2 = 16 - 14 - 2 = 0 \)
L.H.S. = R.H.S. (Verified)

(ii) Multiplication:
\( (x^2 - y)(xy - y^2) = x^2(xy - y^2) - y(xy - y^2) \)
\( = x^3y - x^2y^2 - xy^2 + y^3 \)
Verification:
L.H.S. \( = (1^2 - 2)(1 \times 2 - 2^2) = (1 - 2)(2 - 4) = (-1)(-2) = 2 \)
R.H.S. \( = (1)^3(2) - (1)^2(2)^2 - (1)(2)^2 + (2)^3 = 2 - 4 - 4 + 8 = 2 \)
L.H.S. = R.H.S. (Verified)

(iii) Multiplication:
\( (x - 2y + z)(x - 3z) = x(x - 3z) - 2y(x - 3z) + z(x - 3z) \)
\( = x^2 - 3xz - 2xy + 6yz + zx - 3z^2 = x^2 - 2xz - 2xy + 6yz - 3z^2 \)
Verification:
L.H.S. \( = (-2 - 2(-1) + 1)(-2 - 3(1)) = (1)(-5) = -5 \)
R.H.S. \( = (-2)^2 - 2(-2)(1) - 2(-2)(-1) + 6(-1)(1) - 3(1)^2 = 4 + 4 - 4 - 6 - 3 = -5 \)
L.H.S. = R.H.S. (Verified)
In simple words: First expand the product of the two brackets. Then, show that plugging the numbers into the original expression gives the exact same result as plugging them into your expanded expression.

Exam Tip: Showing L.H.S. = R.H.S. is the best way to prove your multiplication is completely correct.

Question 7. Simplify :
(i) \( 5(x + 3y) - 2(3x - 4y) \)
(ii) \( 3x - 8(5x - 10) \)
(iii) \( 6\{3x - 8(5x - 10)\} \)
(iv) \( 3x - 6\{3x - 8(5x - 10)\} \)
(v) \( 2(3x^2 - 4x - 8) - (3 - 5x - 2x^2) \)
(vi) \( 8x - (3x - \overline{2x - 3}) \)
(vii) \( 12x^2 - (7x - \overline{3x^2 + 15}) \)
Answer:
(i) \( 5(x + 3y) - 2(3x - 4y) = 5x + 15y - 6x + 8y = -x + 23y \)
(ii) \( 3x - 8(5x - 10) = 3x - 40x + 80 = -37x + 80 \)
(iii) \( 6\{3x - 8(5x - 10)\} = 6\{3x - 40x + 80\} = 18x - 240x + 480 = -222x + 480 \)
(iv) \( 3x - 6\{3x - 8(5x - 10)\} = 3x - 6\{3x - 40x + 80\} = 3x - 18x + 240x - 480 = 225x - 480 \)
(v) \( 2(3x^2 - 4x - 8) - (3 - 5x - 2x^2) = 6x^2 - 8x - 16 - 3 + 5x + 2x^2 = 8x^2 - 3x - 19 \)
(vi) \( 8x - (3x - \overline{2x - 3}) = 8x - (3x - 2x + 3) = 8x - 3x + 2x - 3 = 7x - 3 \)
(vii) \( 12x^2 - (7x - \overline{3x^2 + 15}) = 12x^2 - (7x - 3x^2 - 15) = 12x^2 - 7x + 3x^2 + 15 = 15x^2 - 7x + 15 \)
In simple words: Expand the expressions by multiplying and pay close attention to signs when dropping brackets.

Exam Tip: Be very careful when expanding nested brackets, working from the inside out to avoid sign errors.

Question 8. If \( x = -3 \), find the value of : \( 2x^3 + 8x^2 - 15 \).
Answer:
Substituting \( x = -3 \):
\( 2x^3 + 8x^2 - 15 = 2(-3)^3 + 8(-3)^2 - 15 \)
\( = 2(-27) + 8(9) - 15 \)
\( = -54 + 72 - 15 \)
\( = -69 + 72 = 3 \)
In simple words: Replace \( x \) with \( -3 \), compute the cube and square first, then add and subtract the values.

Exam Tip: Remember that negative numbers raised to odd powers yield negative products, while even powers yield positive ones.