Solve by any method except cross multiplication$a(x+y)+b(x-y)=a^2-ab+b^2$ and $a(x+y)-b(x-y)=a^2+ab+b^2$

Given:

$a(x+y)+b(x-y)=a^2-ab+b^2$  ..i)

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

To find: The values of $x$ and $y$. 

Solution

Adding both the equations

$a(x+y)+b(x-y)=a^2-ab+b^2$  ..i)

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

We get, = $a(x + y) + b(x - y) + a(x + y) - b(x - y)$

              = $a^2- ab + b^2 + a^2 +ab - b^2$

             = $x + y = a$ ...iii)

Subtracting ii) from i),

= $a(x + y) + b(x - y) -(a(x + y)  - b(x - y))$

= $a^2 - ab + b^2 - (a^2 + ab -b^2)$

= $ a(x + y) +  b(x - y) - a(x - y) + b(x - y)$

= $a^2 - ab + b^2 - a^2 - ab + b^2$

$\Rightarrow 2b(x - y)$ = $ -2ab + 2b^2$

$\Rightarrow 2b(x - y) = 2b^2 - 2ab$

$\Rightarrow 2b(x - y)$ = $2b(b - a)$

$\Rightarrow x - y = b - a$ ..iv)

Adding iii) & iv),

We get, $\Rightarrow (x + y) + (x - y) a + b- a = 2x = b$

∴ $x = \frac{b}{2}$

Substituting value of x in (iv),

We get, $\Rightarrow x - y = b - a$

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

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

So, the solution is $x = \frac{b}{2}$ and $y= a-\frac{b}{2}$ or $\frac{2a-b}{2}$

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Updated on: 10-Oct-2022

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