Suppose that $x$ $x$ and $y$ $y$ are non zero real numbers such that $\frac{2 x + y}{x - 2 y} = - 3$ $(2x+y)/(x-2y)=-3$ . What is the value of $\frac{2 x^{2} - 4 y + 8}{y^{2} - 2 x + 4}$ $(2x^2-4y+8)/(y^2-2x+4)$ ? A. -1 B. 2 C. 3 D. 4

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Answer 1 · 2018-06-17T09:38:02

The answer is $o p t i o n (B)$ $option (B)$

If $\frac{2 x + y}{x - 2 y} = - 3$ $(2x+y)/(x-2y)=-3$

Then, cross multiply

$2 x + y = - 3 (x - 2 y)$ $2x+y=-3(x-2y)$

$2 x + y = - 3 x + 6 y$ $2x+y=-3x+6y$

$5 x = 5 y$ $5x=5y$

$x = y$ $x=y$

Therefore, as $y = x$ $y=x$

$\frac{2 x^{2} - 4 y + 8}{y^{2} - 2 x + 4}$ $(2x^2-4y+8)/(y^2-2x+4)$ = $\frac{2 (x^{2} - 2 x + 4)}{x^{2} - 2 x + 4}$ $(2(x^2-2x+4))/(x^2-2x+4)$

$(2(cancel(x^2-2x+4)))/cancel(x^2-2x+4)$

$=2$

The answer is $option (B)$

Suppose that xxx and yyy are non zero real numbers such that (2x+y)/(x-2y)=-32x+yx−2y=−3(2x+y)/(x-2y)=-3. What is the value of (2x^2-4y+8)/(y^2-2x+4)2x2−4y+8y2−2x+4(2x^2-4y+8)/(y^2-2x+4)? A. -1 B. 2 C. 3 D. 4