Question

What is the product? startfraction 2 y over y minus 3 endfraction divided by startfraction 4 y minus 12 over 2 y 6 endfraction two-thirds ten-ninths startfraction 4 y over y minus 3 endfraction startfraction 4 y over y 3 endfraction

Answer 1

The product of the expression $\rm \dfrac{\dfrac{2y}{y-3} }{\dfrac{4y-12}{2y+6}}$ is $\rm \dfrac{(y^2+3y)}{(y^2-6y+9)}$.

What is the product?

In mathematics, the term 'product' refers to the answer to a multiplication problem.

The product of the given expression is determined in the following steps given below.

$\rm =\dfrac{\dfrac{2y}{y-3} }{\dfrac{4y-12}{2y+6}}\\\\\\=\dfrac{2y}{y-3}\times \dfrac{2y+6}{4y-12}\\\\= \dfrac{2y \times 2y + 2y \times 6}{y(4y-12)-3(4y-12)}\\\\=\dfrac{4y^2+12y}{4y^2-12y-12y+36}\\\\=\dfrac{4y^2+12y}{4y^2-24y+36}\\\\=\dfrac{4(y^2+3y)}{4(y^2-6y+9)}\\\\ =\dfrac{(y^2+3y)}{(y^2-6y+9)}$

Hence, the product of the expression $\rm \dfrac{\dfrac{2y}{y-3} }{\dfrac{4y-12}{2y+6}}$ is $\rm \dfrac{(y^2+3y)}{(y^2-6y+9)}$.

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