Question

if x^2+y^2-xy=3 and y-x=1 then find ​

Answer 1

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${x}^{2} + {y}^{2} - xy = 3$

Given

$y - x = 1$

$y = 1 + x$

Put y's value for x

${x}^{2} + ( {x + 1})^{2} - x(x + 1) = 3$

${x}^{2} + {x}^{2} + 1 + 2x - {x}^{2} - x = 3$

${x}^{2} + x - 2 = 0$

${x}^{2} + 2x - x - 2$

Take 2x common from first two and (-) from other two.

$2x(x - 2) - 1(x - 2)$

Then x=

$x = \frac{1}{2}$

And

$x = 2$

Put x for y 'svalue

$y = 1 + x$

$y = 1 + \frac{1}{2} = \frac{3}{2}$

And

$y = 3=(2 + 1)$

Ans....

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