Q) Solve the following pair of linear equations:

(a – b) x + (a + b) y = a2 – 2 a b – b2

(a + b) (x + y) = a2 + b2

Ans: 

Given the equations:

(a – b) x + (a + b) y = a2 – 2 a b – b2…… (i)

(a + b) (x + y) = a2 + b2……..(ii)

Step 1: Let’s subtract equation (ii) from equation (i) and we get:

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

∴ [(a – b) x + (a + b) y] – [(a + b) x + (a + b) y)] = [a2 – 2 a b – b2 – a2 – b2]

∴ [(a – b) x + (a + b) y – (a + b) x – (a + b) y)] = [a2 – 2 a b – b2 – a2 – b2]

∴ [(a – b) x – (a + b) x ] = [ – 2 a b – b2  – b2]

∴ [a x – b x – a x – b x] = [- 2 a b – 2 b2 ]

∴ [ – b x – b x] = [- 2 b ( a + b) ]

∴ [- 2 b x ] = [- 2 b ( a + b) ]

∴ x = ( a + b)

Step 2: Let’s substitute x = (a = b) in equation (ii) and we get:

(a + b) ( x + y) = a2 + b2

∴ (a + b) [(a + b) + y] = a2 + b2

∴ (a + b)2 + (a + b) y = a2 + b2

∴ (a + b) y = a2 + b2 – (a + b)2

∴ (a + b) y = a2 + b2 – (a2 + b2 + 2 a b)

∴ (a + b) y = a2 + b2 – a2 –  b2  –  2 a b

∴ (a + b) y =  –  2 a b

∴ y =  \frac{-  2 a b}{(a + b)}

Therefore, x = (a + b) and y =  \frac{-  2 a b}{(a + b)}

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