How do you find the quotient of (5x^2)/(x^2-5x+4)div(10x)/(x-1)?

May 17, 2017

See a solution process below:

Explanation:

First, factor the denominator of the fraction on the left:

$\frac{5 {x}^{2}}{{x}^{2} - 5 x + 4} \div \frac{10 x}{x - 1} \implies \frac{5 {x}^{2}}{\left(x - 4\right) \left(x - 1\right)} \div \frac{10 x}{x - 1}$

Next, rewrite this expression as:

$\frac{\frac{5 {x}^{2}}{\left(x - 4\right) \left(x - 1\right)}}{\frac{10 x}{x - 1}}$

Then, use this rule of dividing fractions to rewrite the expression again and find the quotient:

$\frac{\frac{\textcolor{red}{\left(5 {x}^{2}\right)}}{\textcolor{b l u e}{\left(\left(x - 4\right) \left(x - 1\right)\right)}}}{\frac{\textcolor{g r e e n}{\left(10 x\right)}}{\textcolor{p u r p \le}{\left(x - 1\right)}}} \implies \frac{\textcolor{red}{\left(5 {x}^{2}\right)} \times \textcolor{p u r p \le}{\left(x - 1\right)}}{\textcolor{b l u e}{\left(\left(x - 4\right) \left(x - 1\right)\right)} \times \textcolor{g r e e n}{\left(10 x\right)}}$

$\frac{\textcolor{red}{\left(\cancel{5} \cancel{{x}^{2}} x\right)} \times \cancel{\textcolor{p u r p \le}{\left(x - 1\right)}}}{\textcolor{b l u e}{\left(\left(x - 4\right) \cancel{\left(x - 1\right)}\right)} \times \textcolor{g r e e n}{\left(\cancel{10} 2 \cancel{x}\right)}} \implies$

$\frac{x}{2 \left(x - 4\right)}$ or $\frac{x}{2 x - 8}$