Question

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Answer 1

Question

find the value of a and b if two equation have infinite solution

solution

$so \: incase \: of \: infinite \: many \: solutios \: \\ \\ \frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2} \\ where \: a1 = 1 . \: \: b1 = 2 \: . \: \: c2 = 1$

$and \: a2 = a - b \: . \: \: b2 = a + b \: . \: c1 = a + b - 2 \\ \\ by \: putting \: value \: \\ \\ \frac{1}{a - b} = \frac{2}{a + b} = \frac{1}{a + b - 2} \\ \\ \frac{1}{a - b} = \frac{2}{a + b} \: \: \: \: \: \: also\: \: \: \: \: \: \frac{2}{a + b} = \frac{1}{a + b - 2}$

$bcm \: \\a + b = 2(a - b) \: \: \: also \: \: \: 2(a + b - 2) = a + b \: \\ \\ a + b = 2a - 2b\: \: \: also \: \: \: 2a + 2b - 4 \: = a + b \\ \: \\ a + b = 2a - 2b\: \: \: also \: \: \: 2a + 2b - 4 \: = a + b \\ - 3b = a - - - (2) \: also \: a + b = + 4 - - - (1) \\ by \: putting \: values \: of \: a \: in \: equation \: 1 \\ \\ - 3b + b = 4 \\ - 2b = 4 \\ b = - 2 \\ the \: n \: a \: = value \: of \: b \: in \: equation \: 1 \\ \\ - 3b = a \\ - 3( - 2) = a \\ so \: a = 6$

then value of a = 6 and b = -2

hope it helps you

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