How do I multiply $\frac{8y-4}{10y-5}\times \frac{5y-15}{3y-9}$?

Question

I am not sure how to multiply these fractions. Do I cross multiply or multiply the numerators together and multiply the denominators together? How do I multiply $\frac{8y-4}{10y-5}\times \frac{5y-15}{3y-9}$? It would also help if work is shown.

Answer 1

It is easy as multiplying the numerators and denominators. You would need to multiply both the numerators together and the denominators together: $$\frac {8y - 4}{10y - 5} \cdot \frac {5y - 15}{3y - 9} = \frac {40y^2 - 140y + 60}{30y^2 - 105y + 45}$$ When you simplify the result, you get: $$\frac 43$$ To get this, you have to factor the numerator and denominator and cancel common factors: $$\frac {20(y - 3)(2y - 1)}{15(y - 3)(2y - 1)} = \frac {20}{15} = \frac 43$$

Edit:

A comment just notified me that in order for this to work, "$y$ has to not be equal to 3 or -0.5. The reason (as another comment by the same commenter) is that for $y - 3$, if $y = 3$, $y - 3 = 0$, and you can't divide 0 by itself to get 1.

Answer 2

Edit: I focused in this answer on how to multiply the algebraic expression in the question assuming the OP wants to know the basic multiplication rule. This was triggered by the question header "how to multiply"...Attempting to factor (simplify the expression) is the better technical approach indeed. However, I assumed the OP was after the multiplication procedure basics.

The idea of multiplying similar expressions is: $$(a+b)*(c+d)=a*c+a*d+b*c+b*d$$

So in your case:

$$\frac{8y-4}{10y-5}\times \frac{5y-15}{3y-9}=$$

Note: If you know how to factor the expression, it is a good place to do that here before multiplying, otherwise let's start multiplying...

$$\frac{(8y-4)(5y-15)}{(10y-5)(3y-9)}=$$

$$\frac{8y(5y)+8y(-15)-4(5y)+(-4)(-15)}{ 10y(3y)+10y(-9)+(-5)(3y)+(-5)(-9)}=$$

$$\frac{40y^{2}-120y-20y+60}{ 30y^{2}-90y-15y+45}=$$

$$\frac{40y^{2}-140y+60}{ 30y^{2}-105y+45}$$

You can simplify the above by taking a common factor, however this is easier done in the original form.