Question

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Answer 1

Answer:

Step-by-step explanation:

Given :

a(x + y) + b(x - y) = a² - ab + b²

a(x + y) - b(x - y) = a² + ab + b²

Find x & y,.

Solution :

a(x + y) + b(x - y) = a² - ab + b²

⇒ ax + ay + bx - by = a² - ab + b²

⇒ x(a + b) + y(a - b) = a² - ab + b² ...(i)

a(x + y) - b(x - y) = a² + ab + b²

⇒ ax + ay - bx + by = a² + ab + b²

⇒ x(a - b) + y(a + b) = a² + ab + b² ...(ii)

by adding both the equations,

We get,

⇒ [x(a + b) + y(a - b)] + [x(a - b) + y(a + b)] = [a² - ab + b²] + [a² - ab + b²]

⇒ x [(a + b) + (a - b)] + y [(a + b) + (a - b)] = 2a² + 2b² = 2(a² + b²)

⇒  x(2a) + y(2a) = 2(a² + b²)

⇒ 2a(x + y) = 2(a² + b²)

⇒ x + y = $\frac{2(a^2+b^2)}{2a}$

⇒ $x + y =\frac{a^2+b^2}{a}$ ...(iii)

By substituting the value of (x + y) in given equation :

(given equation)

⇒ a(x + y) + b(x - y) = a² - ab + b²

⇒ $a(\frac{a^2+b^2}{a}) + b(x - y) = a^2 - ab + b^2$

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

⇒ $b(x - y) = -ab$

⇒ b(x - y) = b(-a)

⇒ x - y = -a ...(iv)

By adding both (iii) & (iv),

We get,

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

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

⇒ x = $\frac{b^2}{2a}$

By substituting value of x in (iii),

We get,

⇒ x + y = $\frac{a^2+b^2}{a}$

⇒ $\frac{b^2}{2a} + y = \frac{a^2+b^2}{a}$

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