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y/2y+4 - 3/x+2.
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you got \(x\)'s and \(y\)'s ?
Yes
such as y = x2 or y = 3x + 1
Y 3 -------- MINUS -------- 2y + 4 x+2
\[\frac{y}{2y+4}-\frac{3}{x+2}\]\[=\frac{y(x+2)-3(2y+4)}{(2y+4)(x+2)}\] as a start
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wuts y
then multiply out in the numerator and that will be all
\[=\frac{y(x+2)-3(2y+4)}{(2y+4)(x+2)}\] \[=\frac{xy+2y-6y-12}{(2y+4)(x+2)}=\frac{xy-4y-12}{(2y+4)(x+2)}\]
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