If 4x-y-2x-y-2 then prove that x2-y2-3x-y

4x+y+2x y=2 ⇒ 2(2x+y+1x y)=2 ⇒2x+y+1x y=1 ⇒2x 2y+x+yx^2 y^2=1 ⇒3x yx^2 y^2=1 ⇒x^2 y^2=3x y ...

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Standard VI

Mathematics

Word Problems

If 4x+y+2x-...

Question

If $\frac{4}{x + y} + \frac{2}{x - y} = 2$ then prove that $x^{2} - y^{2} = 3 x - y$

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Solution

$\frac{4}{x + y} + \frac{2}{x - y} = 2$
$\Rightarrow 2 (\frac{2}{x + y} + \frac{1}{x - y}) = 2$
$\Rightarrow \frac{2}{x + y} + \frac{1}{x - y} = 1$
$\Rightarrow \frac{2 x - 2 y + x + y}{x^{2} - y^{2}} = 1$
$\Rightarrow \frac{3 x - y}{x^{2} - y^{2}} = 1$
$\Rightarrow x^{2} - y^{2} = 3 x - y$

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If 4x+y+2x−y=2 then prove that x2−y2=3x−y

4x+y+2x−y=2⇒2(2x+y+1x−y)=2⇒2x+y+1x−y=1⇒2x−2y+x+yx2−y2=1⇒3x−yx2−y2=1⇒x2−y2=3x−y

If $\frac{4}{x + y} + \frac{2}{x - y} = 2$ then prove that $x^{2} - y^{2} = 3 x - y$

$\frac{4}{x + y} + \frac{2}{x - y} = 2$
$\Rightarrow 2 (\frac{2}{x + y} + \frac{1}{x - y}) = 2$
$\Rightarrow \frac{2}{x + y} + \frac{1}{x - y} = 1$
$\Rightarrow \frac{2 x - 2 y + x + y}{x^{2} - y^{2}} = 1$
$\Rightarrow \frac{3 x - y}{x^{2} - y^{2}} = 1$
$\Rightarrow x^{2} - y^{2} = 3 x - y$