If x - y-x - y - z- then y can be written as

The correct option is A x (1 z)/(1 + z) Given equation is: #160;z= x y x+y zx +zy = x y zy+y= x zx y(z+1)= x(1 z) ∴ y=x(1 z) ...

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Byju's Answer

Standard VII

Mathematics

Equation

If x - y/x ...

Question

If $\frac{x - y}{x + y} = z$ , then $y$ can be written as

\frac{x (1 - z)}{(1 + z)}

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\frac{x (1 + 2 z)}{(1 + 2 z)}

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\frac{x (1 - 2 z)}{(1 + z)}

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\frac{x (2 - z)}{(1 + 2 z)}

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Solution

The correct option is A $\frac{x (1 - z)}{(1 + z)}$
Given equation is: $z = \frac{x - y}{x + y}$
$z x + z y = x - y z y + y = x - z x$
$y (z + 1) = x (1 - z)$

$∴ y = \frac{x (1 - z)}{(1 + z)}$

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and

Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} x + y + 2 z & x & y z & y + z + 2 x & y z & x & z + x + 2 y \end{matrix} ∣ ∣ ∣ ∣

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If x−yx+y=z, then y can be written as

The correct option is A x(1−z)(1+z)Given equation is: z=x−yx+yzx+zy=x−yzy+y=x−zxy(z+1)=x(1−z)∴y=x(1−z)(1+z)

If $\frac{x - y}{x + y} = z$ , then $y$ can be written as

The correct option is A $\frac{x (1 - z)}{(1 + z)}$
Given equation is: $z = \frac{x - y}{x + y}$
$z x + z y = x - y z y + y = x - z x$
$y (z + 1) = x (1 - z)$

$∴ y = \frac{x (1 - z)}{(1 + z)}$