Solve the following pairs of linear equations:
(i) $ p x+q y=p-q $
$q x-p y=p+q$
(ii) $ a x+b y=c $
$b x+a y=1+c$
,b>(iii) $ \frac{x}{a}-\frac{y}{b}=0 $
$a x+b y=a^{2}+b^{2}$
(iv) $ (a-b) x+(a+b) y=a^{2}-2 a b-b^{2} $
$(a+b)(x+y)=a^{2}+b^{2}$
(v) $ 152 x-378 y=-74 $
$-378 x+152 y=-604$.

AcademicMathematicsNCERTClass 10

To do:

We have to solve the given pairs of linear equations.

Solution:

(i) $px + qy = p – q$..…(i)

$qx – py = p + q$.…(ii)
Multiplying equation (i) by $p$ and equation (ii) by $q$ and adding the results, we get,

$x(p^2 + q^2) = p(p – q) + q(p + q)$

$x=\frac{p(p-q)}{p^{2}+q^{2}}+\frac{q(p+c_{5}}{p^{2}+q}$

$x=\frac{p^{2}-p q+q p+q^{2}}{p^{2}+q^{2}}$

$x=\frac{p^{2}+q^{2}}{p^{2}+q^{2}}$

$x=1$

Substituting $x=1$ in equation (ii), we get,

$q(1)-p y=p+q$

$q-p y=p+q$

$y=\frac{-p}{p}$

$y=-1$

(ii) $ax + by = c$....…(i)

$bx – ay = 1 + c$.....…(ii)

Multiplying equation (i) by $b$ and equation (ii) by $a$, we get,

$abx + b^2y = cb$....…(iii)

$abx + a^2y = a(1+ c)$…....(iv)

Subtracting (iii) from (iv), we get

$(b^{2}-a^{2}) y=c b-a-a c$

$y=\frac{c(b-a)-a}{b^{2}-a^{2}}$

$y=\frac{c(a-b)+a}{a^{2}-b^{2}}$

$b x +a[\frac{c(a-b)+a}{a^{2}-b^{2}}]=1+c$

$b x=1+c-a[\frac{c(a-b)+a}{a^{2}-b^{2}}]$

$b x=1+c-a[\frac{c a-c b+a}{a^{2}-b^{2}}]$

$bx=\frac{a^{2}-b^{2}+a^{2} c-b^{2} c-a^{2} c+a b c-a^{2}}{a^{2}-b^{2}}$

$bx=\frac{-b^{2}-b^{2} c+a b c}{a^{2}-b^{2}}$

$bx=\frac{b(-b-c b+a c^{2}}{a^{2}-b^{2}}$

$x=\frac{-b-c b+a c}{a^{2}-b^{2}}$

$x=\frac{c(a-b)-b}{a^{2}-b^{2}}$

$x=\frac{c(a-b)-b}{a^{2}-b^{2}}$

Therefore,

$x=\frac{c(a-b)-b}{a^{2}-b^{2}}$ and $y=\frac{c(a-b)+a}{a^{2}-b^{2}}$ is the required solution.

(iii) The given equations can be written as,

$bx - ay = 0$.....…(i)

$ax + by = a^2 + b^2$....…(ii)

Multiplying equation (i) by $b$ and equation (ii) by $a$, we get,

$b^2x - aby = 0$.....….(iii)

$a^2x + aby = a(a^2 + b^2)$…......(iv)

Adding equations (iii) and equation (iv), we get,

$(a^2 + b^2)x = a (a^2+ b^2)$

$x=\frac{a(a^{2}+b^{2})}{a^{2}+b^{2}}$

$x=a$

Substituting $x=a$ in equation (ii), we get,

$a(a)+b y=a^{2}+b^{2}$

$a^2+b y =a^{2}+b^{2}$

$y=\frac{b^2}{b}$

$y=b$

(iv) The given equations can be written as,

\( (a-b) x+(a+b) y=a^{2}-2 a b-b^{2} \).....…(i)

$(a+b)(x+y)=a^{2}+b^{2}$....…(ii)

Subtracting (ii) from (i), we get, we get,

$(a-b)x - (a + b)x = (a^2 - 2ab - b^2) - (a^2 +b^2)$

$x(a - b - a - b) = a^2 - 2ab - b^2 - a^2 - b^2$

$-2bx = -2ab - 2b^2$

$2bx = 2b^2 + 2ab$

$x=\frac{b(a+b)}{b}$

$x=a+b$

Substituting $x=a+b$ in (i), we get,

$(a-b)(a+b)+(a+b) y=a^{2}-2 a b-b^{2}$

$a^{2}-b^{2}+(a+b) y=a^{2}-2 a b-b^{2}$

$(a+b) y =a^{2}-2 a b-b^{2}-a^{2}+b^{2}$

$y=\frac{-2 a b}{a+b}$

(v) The given equations can be written as,

\( 152 x-378 y=-74 \)

$2(76x-189y)=2(-37)$

$76x-189y=-37$.....…(i)

$-378 x+152 y=-604$

$2(-189x+76y)=2(-302)$

$-189x+76y=-302$....…(ii)

Multiplying (i) by 76 and (ii) by 189, we get,

$5776x - 14364y = -2812$...…(iii)

$-35721x + 14364y = -57078$....…(iv)

Adding equations (iii) and (iv), we get,

$5776x - 35721x = -2812 - 57078$

$- 29945x = -59890$

$x = 2$

Substituting $x = 2$ in equation (i), we get,

$76(2)- 189y = -37$

$152 - 189y = -37$

$-189y = -189$

$y = 1$

Therefore, $x = 2$ and $y = 1$ is the required solution.

raja
Updated on 10-Oct-2022 13:20:08

Advertisements