Fundamental Concepts of Algebra — Exercise 12(D)

Find the value of:

$2x + 3y$, when $x = 3$ and $y = 4$.

Answer

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

$\Rightarrow 2x + 3y\\[1em] \Rightarrow 2 \times 3 + 3 \times 4\\[1em] \Rightarrow 6 + 12$

Hence, $\bm{2x + 3y = 18}$.

Find the value of:

$4x - 5y$, when $x = 5$ and $y = 3$.

Answer

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

$\Rightarrow 4x - 5y\\[1em] \Rightarrow 4 \times 5 - 5 \times 3\\[1em] \Rightarrow 20 - 15$

Hence, $\bm{4x - 5y = 5}$.

Find the value of:

$3a - 4b + 5c$, when $a = 10, b = 7$ and $c = 5$.

Answer

Substituting $a = 10, b = 7$ and $c = 5$, we get:

$\Rightarrow 3a - 4b + 5c\\[1em] \Rightarrow 3 \times 10 - 4 \times 7 + 5 \times 5\\[1em] \Rightarrow 30 - 28 + 25$

Hence, $\bm{3a - 4b + 5c = 27}$.

Find the value of:

$8m - 5n - 3p$, when $m = 15, n = -4$ and $p = 20$.

Answer

Substituting $m = 15, n = -4$ and $p = 20$, we get:

$\Rightarrow 8m - 5n - 3p\\[1em] \Rightarrow 8 \times 15 - 5 \times (-4) - 3 \times 20\\[1em] \Rightarrow 120 + 20 - 60$

Hence, $\bm{8m - 5n - 3p = 80}$.

Find the value of:

$5ab + 7bc - 3ca$, when $a = 2, b = 3$ and $c = 4$.

Answer

Substituting $a = 2, b = 3$ and $c = 4$, we get:

$\Rightarrow 5ab + 7bc - 3ca\\[1em] \Rightarrow 5 \times 2 \times 3 + 7 \times 3 \times 4 - 3 \times 4 \times 2\\[1em] \Rightarrow 30 + 84 - 24$

Hence, $\bm{5ab + 7bc - 3ca = 90}$.

If $x = -3$, find the value of:

$5 - 2x$

Answer

Substituting $x = -3$ in the given expression, we get:

$\Rightarrow 5 - 2x\\[1em] \Rightarrow 5 - 2 \times (-3)\\[1em] \Rightarrow 5 + 6$

Hence, $\bm{5 - 2x = 11}$.

If $x = -3$, find the value of:

$x^2 + 2$

Answer

Substituting $x = -3$ in the given expression, we get:

$\Rightarrow x^2 + 2\\[1em] \Rightarrow (-3)^2 + 2\\[1em] \Rightarrow 9 + 2$

Hence, $\bm{x^2 + 2 = 11}$.

If $x = -3$, find the value of:

$6 - x^2$

Answer

Substituting $x = -3$ in the given expression, we get:

$\Rightarrow 6 - x^2\\[1em] \Rightarrow 6 - (-3)^2\\[1em] \Rightarrow 6 - 9$

Hence, $\bm{6 - x^2 = -3}$.

If $x = 5$, find the value of:

$x^2 + x + 3$

Answer

Substituting $x = 5$ in the given expression, we get:

$\Rightarrow x^2 + x + 3\\[1em] \Rightarrow (5)^2 + 5 + 3\\[1em] \Rightarrow 25 + 5 + 3$

Hence, $\bm{x^2 + x + 3 = 33}$.

If $x = 5$, find the value of:

$5 + 4x - x^2$

Answer

Substituting $x = 5$ in the given expression, we get:

$\Rightarrow 5 + 4x - x^2\\[1em] \Rightarrow 5 + 4 \times 5 - (5)^2\\[1em] \Rightarrow 5 + 20 - 25$

Hence, $\bm{5 + 4x - x^2 = 0}$.

If $x = 5$, find the value of:

$9x^2 - 7x$

Answer

Substituting $x = 5$ in the given expression, we get:

$\Rightarrow 9x^2 - 7x\\[1em] \Rightarrow 9 \times (5)^2 - 7 \times 5\\[1em] \Rightarrow 9 \times 25 - 35\\[1em] \Rightarrow 225 - 35$

Hence, $\bm{9x^2 - 7x = 190}$.

Find the value of:

$6x^2 - 5x + 4$, when $x = 5$.

Answer

Substituting $x = 5$, we get:

$\Rightarrow 6x^2 - 5x + 4\\[1em] \Rightarrow 6 \times (5)^2 - 5 \times 5 + 4\\[1em] \Rightarrow 150 - 25 + 4$

Hence, $\bm{6x^2 - 5x + 4 = 129}$.

Find the value of:

$3x^2 - 4x - 2$, when $x = -4$.

Answer

Substituting $x = -4$, we get:

$\Rightarrow 3x^2 - 4x - 2\\[1em] \Rightarrow 3 \times (-4)^2 - 4 \times (-4) - 2\\[1em] \Rightarrow 48 + 16 - 2$

Hence, $\bm{3x^2 - 4x - 2 = 62}$.

Find the value of:

$2x^3 - 3x^2 + 5x + 6$, when $x = 3$.

Answer

Substituting $x = 3$, we get:

$\Rightarrow 2x^3 - 3x^2 + 5x + 6\\[1em] \Rightarrow 2 \times (3)^3 - 3 \times (3)^2 + 5 \times 3 + 6\\[1em] \Rightarrow 54 - 27 + 15 + 6$

Hence, $\bm{2x^3 - 3x^2 + 5x + 6 = 48}$.

Find the value of:

$x^4 - 2x^3 + x^2 - 7x - 10$, when $x = -2$.

Answer

Substituting $x = -2$, we get:

$\Rightarrow x^4 - 2x^3 + x^2 - 7x - 10\\[1em] \Rightarrow (-2)^4 - 2 \times (-2)^3 + (-2)^2 - 7 \times (-2) - 10\\[1em] \Rightarrow 16 + 16 + 4 + 14 - 10$

Hence, $\bm{x^4 - 2x^3 + x^2 - 7x - 10 = 40}$.

If $a = 4$ and $b = 5$, find the value of:

$ab - a^2$

Answer

Substituting $a = 4$ and $b = 5$, we get:

$\Rightarrow ab - a^2\\[1em] \Rightarrow 4 \times 5 - (4)^2\\[1em] \Rightarrow 20 - 16$

Hence, $\bm{ab - a^2 = 4}$.

If $a = 4$ and $b = 5$, find the value of:

$6a^2 - 2ab$

Answer

Substituting $a = 4$ and $b = 5$, we get:

$\Rightarrow 6a^2 - 2ab\\[1em] \Rightarrow 6 \times (4)^2 - 2 \times 4 \times 5\\[1em] \Rightarrow 96 - 40$

Hence, $\bm{6a^2 - 2ab = 56}$.

If $a = 4$ and $b = 5$, find the value of:

$5a^2 - 3b^2$

Answer

Substituting $a = 4$ and $b = 5$, we get:

$\Rightarrow 5a^2 - 3b^2\\[1em] \Rightarrow 5 \times (4)^2 - 3 \times (5)^2\\[1em] \Rightarrow 80 - 75$

Hence, $\bm{5a^2 - 3b^2 = 5}$.

If $a = 4$ and $b = 5$, find the value of:

$4b^2 - 3ab$

Answer

Substituting $a = 4$ and $b = 5$, we get:

$\Rightarrow 4b^2 - 3ab\\[1em] \Rightarrow 4 \times (5)^2 - 3 \times 4 \times 5\\[1em] \Rightarrow 100 - 60$

Hence, $\bm{4b^2 - 3ab = 40}$.

If $x = 4, y = 3$ and $z = -2$, find the value of:

$x - 2y + 3z$

Answer

Substituting $x = 4, y = 3$ and $z = -2$, we get:

$\Rightarrow x - 2y + 3z\\[1em] \Rightarrow 4 - 2 \times 3 + 3 \times (-2)\\[1em] \Rightarrow 4 - 6 - 6$

Hence, $\bm{x - 2y + 3z = -8}$.

If $x = 4, y = 3$ and $z = -2$, find the value of:

$xy + yz + zx$

Answer

Substituting $x = 4, y = 3$ and $z = -2$, we get:

$\Rightarrow xy + yz + zx\\[1em] \Rightarrow 4 \times 3 + 3 \times (-2) + (-2) \times 4\\[1em] \Rightarrow 12 - 6 - 8$

Hence, $\bm{xy + yz + zx = -2}$.

If $x = 4, y = 3$ and $z = -2$, find the value of:

$-2xyz$

Answer

Substituting $x = 4, y = 3$ and $z = -2$, we get:

$\Rightarrow -2xyz\\[1em] \Rightarrow -2 \times 4 \times 3 \times (-2)\\[1em] \Rightarrow 48$

Hence, $\bm{-2xyz = 48}$.

If $p = 3, q = -5$, and $r = 4$, find the value of:

$p^2 - 2qr$

Answer

Substituting $p = 3, q = -5$ and $r = 4$, we get:

$\Rightarrow p^2 - 2qr\\[1em] \Rightarrow (3)^2 - 2 \times (-5) \times 4\\[1em] \Rightarrow 9 + 40$

Hence, $\bm{p^2 - 2qr = 49}$.

If $p = 3, q = -5$, and $r = 4$, find the value of:

$q^2 - 3pr$

Answer

Substituting $p = 3, q = -5$ and $r = 4$, we get:

$\Rightarrow q^2 - 3pr\\[1em] \Rightarrow (-5)^2 - 3 \times 3 \times 4\\[1em] \Rightarrow 25 - 36$

Hence, $\bm{q^2 - 3pr = -11}$.

If $p = 3, q = -5$, and $r = 4$, find the value of:

$r^2 - pq$

Answer

Substituting $p = 3, q = -5$ and $r = 4$, we get:

$\Rightarrow r^2 - pq\\[1em] \Rightarrow (4)^2 - 3 \times (-5)\\[1em] \Rightarrow 16 + 15$

Hence, $\bm{r^2 - pq = 31}$.

If $p = 3, q = -5$, and $r = 4$, find the value of:

$pq + qr + rp$

Answer

Substituting $p = 3, q = -5$ and $r = 4$, we get:

$\Rightarrow pq + qr + rp\\[1em] \Rightarrow 3 \times (-5) + (-5) \times 4 + 4 \times 3\\[1em] \Rightarrow -15 - 20 + 12$

Hence, $\bm{pq + qr + rp = -23}$.

If $p = 3, q = -5$, and $r = 4$, find the value of:

$3pq + 10pr + 2qr$

Answer

Substituting $p = 3, q = -5$ and $r = 4$, we get:

$\Rightarrow 3pq + 10pr + 2qr\\[1em] \Rightarrow 3 \times 3 \times (-5) + 10 \times 3 \times 4 + 2 \times (-5) \times 4\\[1em] \Rightarrow -45 + 120 - 40$

Hence, $\bm{3pq + 10pr + 2qr = 35}$.

If $p = 3, q = -5$, and $r = 4$, find the value of:

$p^2 + q^2 + r^2 - pq - qr - pr$

Answer

Substituting $p = 3, q = -5$ and $r = 4$, we get:

$\Rightarrow p^2 + q^2 + r^2 - pq - qr - pr\\[1em] \Rightarrow (3)^2 + (-5)^2 + (4)^2 - 3 \times (-5) - (-5) \times 4 - 3 \times 4\\[1em] \Rightarrow 9 + 25 + 16 + 15 + 20 - 12$

Hence, $\bm{p^2 + q^2 + r^2 - pq - qr - pr = 73}$.

If $a = 3$, and $b = -4$, find the value of:

$\dfrac{a + b}{a - b}$

Answer

Substituting $a = 3$ and $b = -4$, we get:

$\Rightarrow \dfrac{a + b}{a - b}\\[1em] \Rightarrow \dfrac{3 + (-4)}{3 - (-4)}\\[1em] \Rightarrow \dfrac{-1}{3 + 4}\\[1em] \Rightarrow -\dfrac{1}{7}$

Hence, the required value is $\bm{-\dfrac{1}{7}}$.

If $a = 3$, and $b = -4$, find the value of:

$\dfrac{ab}{a^2 + b^2}$

Answer

Substituting $a = 3$ and $b = -4$, we get:

$\Rightarrow \dfrac{ab}{a^2 + b^2}\\[1em] \Rightarrow \dfrac{3 \times (-4)}{(3)^2 + (-4)^2}\\[1em] \Rightarrow \dfrac{-12}{9 + 16}\\[1em] \Rightarrow -\dfrac{12}{25}$

Hence, the required value is $\bm{-\dfrac{12}{25}}$.

If $a = 3$, and $b = -4$, find the value of:

$\dfrac{a^2 + b^2}{b^2 - a^2}$

Answer

Substituting $a = 3$ and $b = -4$, we get:

$\Rightarrow \dfrac{a^2 + b^2}{b^2 - a^2}\\[1em] \Rightarrow \dfrac{(3)^2 + (-4)^2}{(-4)^2 - (3)^2}\\[1em] \Rightarrow \dfrac{9 + 16}{16 - 9}\\[1em] \Rightarrow \dfrac{25}{7}$

Hence, the required value is $\bm{\dfrac{25}{7}}$.

If $a = 8, b = 2$ and $c = 3$, find the value of $\dfrac{16bc - 5ab}{abc - 3b^2}$.

Answer

Substituting $a = 8, b = 2$ and $c = 3$, we get:

$\Rightarrow \dfrac{16bc - 5ab}{abc - 3b^2}\\[1em] \Rightarrow \dfrac{16 \times 2 \times 3 - 5 \times 8 \times 2}{8 \times 2 \times 3 - 3 \times (2)^2}\\[1em] \Rightarrow \dfrac{96 - 80}{48 - 12}\\[1em] \Rightarrow \dfrac{16}{36}\\[1em] \Rightarrow \dfrac{4}{9}$

Hence, the required value is $\bm{\dfrac{4}{9}}$.

If $x = 4, y = 3$ and $z = -2$, find the value of $\dfrac{x^2 + y^2 - z^2}{xy + yz - zx}$.

Answer

Substituting $x = 4, y = 3$ and $z = -2$, we get:

$\Rightarrow \dfrac{x^2 + y^2 - z^2}{xy + yz - zx}\\[1em] \Rightarrow \dfrac{(4)^2 + (3)^2 - (-2)^2}{4 \times 3 + 3 \times (-2) - (-2) \times 4}\\[1em] \Rightarrow \dfrac{16 + 9 - 4}{12 - 6 + 8}\\[1em] \Rightarrow \dfrac{21}{14}\\[1em] \Rightarrow \dfrac{3}{2}$

Hence, the required value is $\bm{\dfrac{3}{2}}$.

State true or false:

(i) $x^2 + y^2 = z^2$, when $x = 5, y = 12$ and $z = 13$.

(ii) The value of $5x - 3$ is 2, when $x = 0$.

(iii) The value of $x^y$ is 8, when $x = 3$ and $y = 2$.

(iv) When $a = 3$, then $a^2$ and $2a$ have equal values.

Answer

(i) True

Reason :

L.H.S. $= x^2 + y^2 = (5)^2 + (12)^2 = 25 + 144 = 169$

R.H.S. $= z^2 = (13)^2 = 169$

Since L.H.S. = R.H.S., the statement is true.

(ii) False

Reason :

At $x = 0, 5x - 3 = 5 \times 0 - 3 = -3$, which is not 2.

(iii) False

Reason :

At $x = 3$ and $y = 2, x^y = 3^2 = 9$, which is not 8.

(iv) False.

Reason :

At $a = 3, a^2 = 9$ and $2a = 6$, which are not equal.