Question

10/x+y+2/x-y=4,15/x+y-5/x-y=-2.by elimination method or subtitution method>3

Answer 1

Solution:-

equation

$\rm \to \: \dfrac{10}{x + y} + \dfrac{2}{x - y} = 4 \: \: \: \: \: ...(i)eq$

$\rm \to \: \dfrac{15}{x + y} - \dfrac{5}{x - y} = - 2 \: \: \: \: \: ....(ii)eq$

let

$\rm \to \: \dfrac{1}{x + y} = u \: \: and \: \: \dfrac{1}{x - y} = v$

we can write as

$\rm \to10u + 2v = 4 \: \: \: \: ....(i)$

$\rm \to15u - 5v = - 2 \: \: \: \: ...(ii)eq$

$\rm \: 10u + 2v = 4 \: \: \: \: ....( \times - 5)\\ \rm \: 15u - 5v = - 2 \: \: \: \: .......( \times 2)$

We get

$\rm \: - 50u - 10v = - 20 \: \: \: ...(i) \\ \rm30u - 10v = - 4 \: \: \: \: \: \: \: \: ....(ii)$

Now sub the equation

$\rm \to \: - 50u - 10v - 30u + 10v = - 20 + 4$

$\rm \to - 80u = - 16$

$\rm \to \: u = \dfrac{1}{5}$

Now put the value of u on( ii)nd eq

$\rm \to15u - 5v = - 2 \: \: \: \: ...(ii)eq$

$\rm \to15 \times \dfrac{1}{5} - 5v = - 2 \: \: \: \: ...(ii)eq$

$\rm \to \: 3 - 5v = - 2$

$\rm \to \: - 5v = - 5$

$\rm \to \: v = 1$

Now take

$\rm \to \: \dfrac{1}{x + y} = u \: \: and \: \: \dfrac{1}{x - y} = v$

By putting the value of u and v we get

$\rm \to \: \dfrac{1}{x + y} = \dfrac{1}{5} \: \: and \: \: \dfrac{1}{x - y} = 1$

Now we get

$\rm \: x + y = 5 \: \: \: \: ...(i)$

$\rm \: x - y = 1 \: \: \: \: \: ...(ii)$

Subtract the equation

$\rm \: x + y - x + y = 5 - 1$

$\rm \to2y = 4$

$\rm \to \: y = 2$

Put the value of y on ii eq

$\rm \to \: x - y = 1$

$\rm \: x - 2 = 1$

$\rm \: x = 3$

So value of x = 3 and y = 2