Framing Algebraic Expressions — Exercise 13(B)

Fill in the following blanks, when : $x = 3, y = 6, z = 18, a = 2, b = 8, c = 32$ and $d = 0$.

(i) $x + y$ = 3 + 6 = 9

(ii) $y - x$ = ......

(iii) $\dfrac{y}{x}$ = ......

(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) $\dfrac{y}{x} = \dfrac{6}{3}$ = 2

(iv) $c \div b$ = 32 ÷ 8 = 4

(v) $z \div x$ = 18 ÷ 3 = 6

(vi) $y \times d$ = 6 × 0 = 0

(vii) $d \div x$ = 0 ÷ 3 = 0

(viii) $ab + y$ = (2 × 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 = 0

(xii) $z - a - b$ = 18 − 2 − 8 = 8

(xiii) $d - a + x$ = 0 − 2 + 3 = 1

(xiv) $xy - bd$ = (3 × 6) − (8 × 0) = 18 − 0 = 18

(xv) $xz + cd$ = (3 × 18) + (32 × 0) = 54 + 0 = 54

Find the value of :

$p + 2q + 3r$, when $p = 1, q = 5$ and $r = 2$

Answer

Substituting $p = 1, q = 5$ and $r = 2$, we get :

$\Rightarrow p + 2q + 3r \\[1em] \Rightarrow 1 + 2(5) + 3(2) \\[1em] \Rightarrow 1 + 10 + 6 \\[1em] \Rightarrow 17.$

Hence, $\bm{p + 2q + 3r = 17}$.

Find the value of :

$2a + 4b + 5c$, when $a = 5, b = 10$ and $c = 20$

Answer

Substituting $a = 5, b = 10$ and $c = 20$, we get :

$\Rightarrow 2a + 4b + 5c \\[1em] \Rightarrow 2(5) + 4(10) + 5(20) \\[1em] \Rightarrow 10 + 40 + 100 \\[1em] \Rightarrow 150.$

Hence, $\bm{2a + 4b + 5c = 150}$.

Find the value of :

$3a - 2b$, when $a = 8$ and $b = 10$

Answer

Substituting $a = 8$ and $b = 10$, we get :

$\Rightarrow 3a - 2b \\[1em] \Rightarrow 3(8) - 2(10) \\[1em] \Rightarrow 24 - 20 \\[1em] \Rightarrow 4.$

Hence, $\bm{3a - 2b = 4}$.

Find the value of :

$5x + 3y - 6z$, when $x = 3, y = 5$ and $z = 4$

Answer

Substituting $x = 3, y = 5$ and $z = 4$, we get :

$\Rightarrow 5x + 3y - 6z \\[1em] \Rightarrow 5(3) + 3(5) - 6(4) \\[1em] \Rightarrow 15 + 15 - 24 \\[1em] \Rightarrow 6.$

Hence, $\bm{5x + 3y - 6z = 6}$.

Find the value of :

$2p - 3q + 4r - 8s$, when $p = 10, q = 8, r = 6$ and $s = 2$

Answer

Substituting $p = 10, q = 8, r = 6$ and $s = 2$, we get :

$\Rightarrow 2p - 3q + 4r - 8s \\[1em] \Rightarrow 2(10) - 3(8) + 4(6) - 8(2) \\[1em] \Rightarrow 20 - 24 + 24 - 16 \\[1em] \Rightarrow 4.$

Hence, $\bm{2p - 3q + 4r - 8s = 4}$.

Find the value of :

$6m - 2n - 5p - 3q$, when $m = 20, n = 10, p = 2$ and $q = 9$

Answer

Substituting $m = 20, n = 10, p = 2$ and $q = 9$, we get :

$\Rightarrow 6m - 2n - 5p - 3q \\[1em] \Rightarrow 6(20) - 2(10) - 5(2) - 3(9) \\[1em] \Rightarrow 120 - 20 - 10 - 27 \\[1em] \Rightarrow 63.$

Hence, $\bm{6m - 2n - 5p - 3q = 63}$.

Find the value of :

$4pq \times 2r$, when $p = 5, q = 3$ and $r = \dfrac{1}{2}$

Answer

Substituting $p = 5, q = 3$ and $r = \dfrac{1}{2}$, we get :

$\Rightarrow 4pq \times 2r \\[1em] \Rightarrow 8pqr \\[1em] \Rightarrow 8 \times 5 \times 3 \times \dfrac{1}{2} \\[1em] \Rightarrow 60.$

Hence, $\bm{4pq \times 2r = 60}$.

Find the value of :

$\dfrac{yx}{z}$, when $x = 8, y = 4$ and $z = 16$

Answer

Substituting $x = 8, y = 4$ and $z = 16$, we get :

$\Rightarrow \dfrac{yx}{z} \\[1em] \Rightarrow \dfrac{4 \times 8}{16} \\[1em] \Rightarrow \dfrac{32}{16} \\[1em] \Rightarrow 2.$

Hence, $\bm{\dfrac{yx}{z} = 2}$.

Find the value of :

$\dfrac{a + b - c}{2a}$, when $a = 5, b = 7$ and $c = 2$

Answer

Substituting $a = 5, b = 7$ and $c = 2$, we get :

$\Rightarrow \dfrac{a + b - c}{2a} \\[1em] \Rightarrow \dfrac{5 + 7 - 2}{2 \times 5} \\[1em] \Rightarrow \dfrac{10}{10} \\[1em] \Rightarrow 1.$

Hence, $\bm{\dfrac{a + b - c}{2a} = 1}$.

If $a = 3, b = 0, c = 2$ and $d = 1$, find the value of :

$3a + 2b - 6c + 4d$

Answer

Substituting $a = 3, b = 0, c = 2$ and $d = 1$, we get :

$\Rightarrow 3a + 2b - 6c + 4d \\[1em] \Rightarrow 3(3) + 2(0) - 6(2) + 4(1) \\[1em] \Rightarrow 9 + 0 - 12 + 4 \\[1em] \Rightarrow 1.$

Hence, $\bm{3a + 2b - 6c + 4d = 1}$.

If $a = 3, b = 0, c = 2$ and $d = 1$, find the value of :

$6a - 3b - 4c - 2d$

Answer

Substituting $a = 3, b = 0, c = 2$ and $d = 1$, we get :

$\Rightarrow 6a - 3b - 4c - 2d \\[1em] \Rightarrow 6(3) - 3(0) - 4(2) - 2(1) \\[1em] \Rightarrow 18 - 0 - 8 - 2 \\[1em] \Rightarrow 8.$

Hence, $\bm{6a - 3b - 4c - 2d = 8}$.

If $a = 3, b = 0, c = 2$ and $d = 1$, find the value of :

$ab - bc + cd - da$

Answer

Substituting $a = 3, b = 0, c = 2$ and $d = 1$, we get :

$\Rightarrow ab - bc + cd - da \\[1em] \Rightarrow (3)(0) - (0)(2) + (2)(1) - (1)(3) \\[1em] \Rightarrow 0 - 0 + 2 - 3 \\[1em] \Rightarrow -1.$

Hence, $\bm{ab - bc + cd - da = -1}$.

If $a = 3, b = 0, c = 2$ and $d = 1$, find the value of :

$abc - bcd + cda$

Answer

Substituting $a = 3, b = 0, c = 2$ and $d = 1$, we get :

$\Rightarrow abc - bcd + cda \\[1em] \Rightarrow (3)(0)(2) - (0)(2)(1) + (2)(1)(3) \\[1em] \Rightarrow 0 - 0 + 6 \\[1em] \Rightarrow 6.$

Hence, $\bm{abc - bcd + cda = 6}$.

If $a = 3, b = 0, c = 2$ and $d = 1$, find the value of :

$a^2 + 2b^2 - 3c^2$

Answer

Substituting $a = 3, b = 0, c = 2$ and $d = 1$, we get :

$\Rightarrow a^{2} + 2b^{2} - 3c^{2} \\[1em] \Rightarrow (3)^{2} + 2(0)^{2} - 3(2)^{2} \\[1em] \Rightarrow 9 + 0 - 12 \\[1em] \Rightarrow -3.$

Hence, $\bm{a^2 + 2b^2 - 3c^2 = -3}$.

If $a = 3, b = 0, c = 2$ and $d = 1$, find the value of :

$a^2 + b^2 - c^2 + d^2$

Answer

Substituting $a = 3, b = 0, c = 2$ and $d = 1$, we get :

$\Rightarrow a^{2} + b^{2} - c^{2} + d^{2} \\[1em] \Rightarrow (3)^{2} + (0)^{2} - (2)^{2} + (1)^{2} \\[1em] \Rightarrow 9 + 0 - 4 + 1 \\[1em] \Rightarrow 6.$

Hence, $\bm{a^2 + b^2 - c^2 + d^2 = 6}$.

If $a = 3, b = 0, c = 2$ and $d = 1$, find the value of :

$2a^2 - 3b^2 + 4c^2 - 5d^2$

Answer

Substituting $a = 3, b = 0, c = 2$ and $d = 1$, we get :

$\Rightarrow 2a^{2} - 3b^{2} + 4c^{2} - 5d^{2} \\[1em] \Rightarrow 2(3)^{2} - 3(0)^{2} + 4(2)^{2} - 5(1)^{2} \\[1em] \Rightarrow 2(9) - 0 + 4(4) - 5(1) \\[1em] \Rightarrow 18 + 16 - 5 \\[1em] \Rightarrow 29.$

Hence, $\bm{2a^2 - 3b^2 + 4c^2 - 5d^2 = 29}$.

Find the value of : $5x^2 - 3x + 2$, when $x = 2$.

Answer

Substituting $x$ = 2, we get :

$\Rightarrow 5x^{2} - 3x + 2 \\[1em] \Rightarrow 5(2)^{2} - 3(2) + 2 \\[1em] \Rightarrow 5(4) - 6 + 2 \\[1em] \Rightarrow 20 - 6 + 2 \\[1em] \Rightarrow 16.$

Hence, $\bm{5x^2 - 3x + 2 = 16}$.

Find the value of : $3x^3 - 4x^2 + 5x - 6$, when $x = -1$.

Answer

Substituting $x$ = −1, we get :

$\Rightarrow 3x^{3} - 4x^{2} + 5x - 6 \\[1em] \Rightarrow 3(-1)^{3} - 4(-1)^{2} + 5(-1) - 6 \\[1em] \Rightarrow 3(-1) - 4(1) - 5 - 6 \\[1em] \Rightarrow -3 - 4 - 5 - 6 \\[1em] \Rightarrow -18.$

Hence, $\bm{3x^3 - 4x^2 + 5x - 6 = -18}$.

Show that the value of : $x^3 - 8x^2 + 12x - 5$ is zero, when $x = 1$.

Answer

Substituting $x$ = 1, we get :

$\Rightarrow x^{3} - 8x^{2} + 12x - 5 \\[1em] \Rightarrow (1)^{3} - 8(1)^{2} + 12(1) - 5 \\[1em] \Rightarrow 1 - 8 + 12 - 5 \\[1em] \Rightarrow 0.$

Hence, the value of $\bm{x^3 - 8x^2 + 12x - 5}$ is zero, when $\bm{x = 1}$.

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) $\dfrac{2x - 4}{x + 1} = -1$, when $x = 1$.

Answer

(i) Substituting $x$ = 1 in $x$ + 5,

⇒ $x$ + 5 = 1 + 5 = 6

Since the value obtained is 6, the statement is true.

(ii) Substituting $x$ = 0 in 2$x$ − 3,

⇒ 2$x$ − 3 = 2(0) − 3 = −3

Since the value obtained is −3 (not 1), the statement is false.

(iii) Substituting $x$ = 1 in $\dfrac{2x - 4}{x + 1}$,

$\Rightarrow \dfrac{2x - 4}{x + 1} = \dfrac{2(1) - 4}{1 + 1} \\[1em] \Rightarrow \dfrac{2 - 4}{2} = \dfrac{-2}{2} = -1.$

Since the value obtained is −1, the statement is true.

If $x = 2, y = 5$ and $z = 4$, find the value of each of the following :

(i) $\dfrac{x}{2x^2}$

(ii) $\dfrac{xz}{yz}$

(iii) $z^x$

(iv) $y^x$

(v) $\dfrac{x^2 y^2 z^2}{xz}$

(vi) $\dfrac{5x^4 y^2 z^2}{2x^2}$

(vii) $xy \div y^2z$

(viii) $\dfrac{x^2 y^x}{x}$

Answer

(i) Substituting $x$ = 2, we get :

$\Rightarrow \dfrac{x}{2x^{2}} = \dfrac{2}{2(2)^{2}} \\[1em] \Rightarrow \dfrac{2}{2 \times 4} = \dfrac{2}{8} = \dfrac{1}{4}.$

Hence, $\bm{\dfrac{x}{2x^{2}} = \dfrac{1}{4}}$.

(ii) Substituting $x$ = 2, $y$ = 5 and $z$ = 4, we get :

$\Rightarrow \dfrac{xz}{yz} = \dfrac{2 \times 4}{5 \times 4} \\[1em] \Rightarrow \dfrac{8}{20} = \dfrac{2}{5}.$

Hence, $\bm{\dfrac{xz}{yz} = \dfrac{2}{5}}$.

(iii) Substituting $z$ = 4 and $x$ = 2, we get :

$\Rightarrow z^{x} = 4^{2} = 16.$

Hence, $\bm{z^x = 16}$.

(iv) Substituting $y$ = 5 and $x$ = 2, we get :

$\Rightarrow y^{x} = 5^{2} = 25.$

Hence, y^x = 25.

(v) Substituting $x$ = 2, $y$ = 5 and $z$ = 4, we get :

$\Rightarrow \dfrac{x^{2} , y^{2} , z^{2}}{xz} = x , y^{2} , z \\[1em] \Rightarrow 2 \times (5)^{2} \times 4 \\[1em] \Rightarrow 2 \times 25 \times 4 = 200.$

Hence, $\bm{\dfrac{x^{2} , y^{2} , z^{2}}{xz} = 200}$.

(vi) Substituting $x$ = 2, $y$ = 5 and $z$ = 4, we get :

$\Rightarrow \dfrac{5x^{4} , y^{2} , z^{2}}{2x^{2}} = \dfrac{5}{2} , x^{2} , y^{2} , z^{2} \\[1em] \Rightarrow \dfrac{5}{2} \times (2)^{2} \times (5)^{2} \times (4)^{2} \\[1em] \Rightarrow \dfrac{5}{2} \times 4 \times 25 \times 16 \\[1em] \Rightarrow 4000.$

Hence, $\bm{\dfrac{5x^{4} , y^{2} , z^{2}}{2x^{2}} = 4000}$.

(vii) Substituting $x$ = 2, $y$ = 5 and $z$ = 4, we get :

$\Rightarrow xy \div y^{2}z = \dfrac{xy}{y^{2}z} = \dfrac{x}{yz} \\[1em] \Rightarrow \dfrac{2}{5 \times 4} = \dfrac{2}{20} = \dfrac{1}{10}.$

Hence, $\bm{xy \div y^2z = \dfrac{1}{10}}$.

(viii) Substituting $x$ = 2 and $y$ = 5, we get :

$\Rightarrow \dfrac{x^{2} , y^{x}}{x} = x , y^{x} \\[1em] \Rightarrow 2 \times (5)^{2} \\[1em] \Rightarrow 2 \times 25 = 50.$

Hence, $\bm{\dfrac{x^{2} , y^{x}}{x} = 50}$.

If $a = 3$, find the values of $a^2$ and $2^a$.

Answer

Substituting $a$ = 3, we get :

$\Rightarrow a^{2} = (3)^{2} = 9 \\[1em] \Rightarrow 2^{a} = 2^{3} = 8.$

Hence, $\bm{a^2 = 9 \text{ and } 2^a = 8}$.

If $m = 2$, find the difference between the values of $4m^3$ and $3m^4$.

Answer

Substituting $m$ = 2, we get :

$\Rightarrow 4m^{3} = 4(2)^{3} = 4 \times 8 = 32 \\[1em] \Rightarrow 3m^{4} = 3(2)^{4} = 3 \times 16 = 48.$

∴ Difference = $3m^4 - 4m^3$ = 48 − 32 = 16

Hence, the required difference is 16.