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How do you simplify $\frac{2 y - 6}{3 y^{2} - y^{3}}$ $(2y-6)/(3y^2-y^3)$ ?

2 Answers

Dec 22, 2016

$- \frac{2}{y^{2}}$ $color(green)(-2/y^2)$

Explanation:

Note that $3 y^{2} - y^{3} = - y^{3} + 3 y^{2}$ $3y^2-y^3=-y^3+3y^2$

$XXX = (- y^{2}) (y - 3)$ $color(white)("XXX")=(-y^2)(y-3)$

$XXX = (- y^{2}) (\frac{(2 y - 6)}{2})$ $color(white)("XXX")=(-y^2)(((2y-6))/2)$

$XXX = (- \frac{y^{2}}{2}) (2 y - 6)$ $color(white)("XXX")=(-(y^2)/2)(2y-6)$

Therefore
$color(white)("XXX")(2y-6)/(-y^3+3y^2)=(cancel(2y-6))/(-y^2/2(cancel(2y-6)))=-2/y^2$

Dec 22, 2016

$=-2/y^2$

Explanation:

$(2y-6)/(3y^2-y^3)" "$ factorise as far as possible

$=(2(y-3))/(y^2(3-y))$

Notice that the brackets are the same except for the signs.
Do a switch round:

$color(red)((3-y) = -(-3+y) = -(y-3))$

$=(2(y-3))/(y^2color(red)((3-y))$

$=(2(y-3))/(-y^2color(red)((y-3))$

$=(2cancel((y-3)))/(-y^2cancel((y-3))$

$=-2/y^2$

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