**Q) Solve the following pair of linear equations:**

**(a – b) x + (a + b) y = a ^{2} – 2 a b – **

**b**^{2}**(a + b) (x + y) = a ^{2} + b^{2}**

**Ans: **

Given the equations:

(a – b) x + (a + b) y = a^{2} – 2 a b – b^{2}…… (i)

(a + b) (x + y) = a^{2} + b^{2}……..(ii)

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

[(a – b) x + (a + b) y] – [(a + b) (x + y)] = [ a^{2} – 2 a b – b^{2 }] – [ a^{2} + b^{2 }]

∴ [(a – b) x + (a + b) y] – [(a + b) x + (a + b) y)] = [a^{2} – 2 a b – b^{2} – a^{2} – b^{2}]

∴ [(a – b) x + (a + b) y – (a + b) x – (a + b) y)] = [a^{2} – 2 a b – b^{2} – a^{2} – b^{2}]

∴ [(a – b) x – (a + b) x ] = [ – 2 a b – b^{2} – b^{2}]

∴ [a x – b x – a x – b x] = [- 2 a b – 2 b^{2} ]

∴ [ – 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) = a^{2 }+ b^{2}

∴ (a + b) [(a + b) + y] = a^{2 }+ b^{2}

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

∴ (a + b) y = a^{2 }+ b^{2 }– (a + b)^{2}

∴ (a + b) y = a^{2 }+ b^{2 }– (a^{2 }+ b^{2 }+ 2 a b)

∴ (a + b) y = a^{2 }+ b^{2 }– a^{2 }– b^{2 } – 2 a b

∴ (a + b) y = – 2 a b

**∴ y = **

**Therefore, x = (a + b) and y = **

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