How do you solve $\frac{1 + 2 y}{y - 4} = \frac{4 y^{2} + 5 y}{2 y^{2} - 7 y - 4}$ $(1+2y)/(y-4 )= (4y^2+5y)/(2y^2-7y-4)$ ?

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Answer 1 · 2015-05-29T21:37:31

$\frac{1 + 2 y}{y - 4} = \frac{4 y^{2} + 5 y}{2 y^{2} - 7 y - 4}$ $(1+2y)/(y−4)=(4y^2+5y)/(2y^2−7y−4)$

$\frac{4 y^{2} + 5 y}{2 y^{2} - 8 y + y - 4}$ $(4y^2+5y)/color(red)(2y^2-8y+y-4)$

$\frac{4 y^{2} + 5 y}{2 y (y - 4) + 1 (y - 4)}$ $(4y^2+5y)/[color(red)(2y(y-4)+1(y-4)]$

$= \frac{4 y^{2} + 5 y}{(1 + 2 y) (y - 4)}$ $=(4y^2+5y)/[(1+2y)(y-4)]$

$(1+2y)/color(red)cancel(y−4)=(4y^2+5y)/[(1+2y)color(red)cancel((y-4)]$

$color(red)[(1+2y)^2]=4y^2+5y$

By expansion: $->(a+b)^2=a^2+2ab+b^2$

$color(red)[1+4y+cancel(4y^2)]=cancel(4y^2)+5y$

ANSWER
$y=1$

Answer 2 · 2015-05-29T21:37:33

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How do you solve (1+2y)/(y-4 )= (4y^2+5y)/(2y^2-7y-4)1+2yy−4=4y2+5y2y2−7y−4(1+2y)/(y-4 )= (4y^2+5y)/(2y^2-7y-4)?