Question

linear equation....... ​

Answer 1

Answer:

$for \: x = 0 \\ y = (a + b) ....(1)\\ and \: for \: y = 0 \\ x = (a + b).....(2) \\ substitute \: (1) \: and \: (2) \: in \: given \: equation \: of \: lhs \: we \: will \: get \\ a(a + b) - b(a + b) = a {}^{2} + ab - ab - b {}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: a {}^{2} - b {}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = rhs \: proved$

Answer 2

Step-by-step explanation:

we take,

x = 0

then,

x + y = a + b

0 + y = a + b

y = a + b .... (I)

Now,

we take y = 0

then,

x + y = a + b

x + 0 = a + b

x = a + b... (ii)

now, putting the value of equation (I) and (ii) respectively:-

=>ax - by

=>a(a+b) - b(a +b)

=> a² + ab -ab - b²

=> a²- b²

L.H.S = R.H.S proved

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