1. If ax + by = 6, bx - ay = 2 and x^2 + y^2 = 4, then the value of (a^2 + b^2) would be:
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By: anil on 05 May 2019 01.59 am
it is given that ax + by = 6..........(1) bx - ay = 2........(2) and x^2 + y^2 = 4
now multiply 1 and 2nd equation by a and b respectively we get $$a^2$$x + $$ab$$y = 6a $$b^2$$x - aby = 2b adding above equations we get, $$a^2 + b^2$$ x = 6a +2b x = $$frac{6a+2b}{a^2 + b^2}$$ Similarly , we get y = $$frac{6a-2b}{a^2 + b^2}$$ putting above values in x^2 + y^2 = 4 we get , a = 1 and b = 3 hence $$1^2 + 3^2$$ = 10
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now multiply 1 and 2nd equation by a and b respectively we get $$a^2$$x + $$ab$$y = 6a $$b^2$$x - aby = 2b adding above equations we get, $$a^2 + b^2$$ x = 6a +2b x = $$frac{6a+2b}{a^2 + b^2}$$ Similarly , we get y = $$frac{6a-2b}{a^2 + b^2}$$ putting above values in x^2 + y^2 = 4 we get , a = 1 and b = 3 hence $$1^2 + 3^2$$ = 10