Answer
$\frac{4}{3}$
Restriction: $y \ne \frac{1}{2}, 3$
Work Step by Step
Let's first take a look at the problem to see what we can cancel out from the numerator and denominator:
$\dfrac{8y - 4}{10y - 5} \cdot \dfrac{5y - 15}{3y - 9}$
Factor out terms from the binomials:
$\dfrac{4(2y - 1)}{5(2y - 1)} \cdot \dfrac{5(y - 3)}{3(y - 3)}$
Multiply fractions:
$\dfrac{20(2y - 1)(y - 3)}{15(2y - 1)(y - 3)}$
Divide numerator and denominator by $5$ to simplify:
$\dfrac{4(2y - 1)(y - 3)}{3(2y - 1)(y - 3)}$
Cancel out the factors $2y - 1$ and $y - 3$:
$\dfrac{4}{3}$
To find out if there are any restrictions on the variables, we need to find which values of $y$ will cause the denominator to equal zero, which would make the whole expression undefined.
Set each factor equal to zero, according to the zero product property.
First factor:
$10y - 5 = 0$
Add $5$ to each side of the equation:
$10y = 5$
Divide each side by $10$:
$y = \frac{5}{10}$
Simplify the fraction:
$y = \frac{1}{2}$
Second factor:
$3y - 9 = 0$
Add $9$ to each side of the equation:
$3y = 9$
Divide each side by $3$:
$y = \frac{9}{3}$
Simplify the fraction:
$y = 3$
Restriction:
$y \ne \frac{1}{2}, 3$
