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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.

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    $\begingroup$ Multiplication of fractions is a case of multiplying numerators and denominators. $\endgroup$ – ÍgjøgnumMeg Mar 18 '16 at 0:44
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    $\begingroup$ But first you should completely factor each numerator and each denominator and check for any factors which might be common to both the numerator and the denominator. $\endgroup$ – John Wayland Bales Mar 18 '16 at 0:54
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    $\begingroup$ You can use "cross multiplication" when you are trying to solve an equation between two fractions. This is equivalent to multiplying both sides of an equation by the common denominator. But in this problem, there is no equation, you are merely simplifying an algebraic expression. $\endgroup$ – John Wayland Bales Mar 18 '16 at 0:57
  • $\begingroup$ Okay thank you. And say if the multiplication sign was replaced by a division sign. What would I do then? $\endgroup$ – Muniza Siddiqui Mar 18 '16 at 1:01
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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.

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  • $\begingroup$ Your answer is clearly better than mine. I assumed that the OP does not know how to factor since he/she was asking about multiplication... $\endgroup$ – NoChance Mar 18 '16 at 1:08
  • $\begingroup$ @NoChance, that is a thoughtful compliment, but really, you were working hard enough to answer the question. Your answer also showed some work. Your answer was as well as mine. $\endgroup$ – Obinna Nwakwue Mar 18 '16 at 1:10
  • $\begingroup$ Thank you. Maybe you want to add the fact that to cancel out the expressions, y should not be equal to $3$ or $-0.5$, otherwise the denominator would be zero and we can't cancel like expressions. $\endgroup$ – NoChance Mar 18 '16 at 1:12
  • $\begingroup$ Well, that may be true (if you have evidence for it), but when it comes to canceling factors, you don't need to know the value of $y$. It doesn't matter if $y = 3, 0.5, \text {etc.}$. You just have to find common factors and cancel. $\endgroup$ – Obinna Nwakwue Mar 18 '16 at 1:14
  • $\begingroup$ Actually it does make a difference. For example, if $y=3$, the expression $y-3$ becomes zero, and we can't divide $0$ by $0$ to get $1$. The same goes for $x=-0.5$. $\endgroup$ – NoChance Mar 18 '16 at 1:18
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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.

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  • $\begingroup$ Okay thank you. And say if the multiplication sign was replaced by a division sign. What would I do then? $\endgroup$ – Muniza Siddiqui Mar 18 '16 at 1:05
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    $\begingroup$ Division requires factoring. the expression. This is not trivial. However, take a look at the answer given by @Obinna Nwakwue. $\endgroup$ – NoChance Mar 18 '16 at 1:07
  • $\begingroup$ It is better ti FIRST factor and cancel common factors and THEN multiply the resulting factors rather than the other way around, otherwise simplifying can become quite cumbersome. I did not put any downvote, but I will readily admit that both provided answers are, as far as the preferred method is concerned, very poor... $\endgroup$ – imranfat Mar 18 '16 at 2:41
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    $\begingroup$ @imranfat ,Your comment regarding factoring first is correct. My assumption when I read the question "How To Multiply..." lead me to think that the OP is not familiar with multiplication. I did not expect the OP to know how to factor if he/she did not know how to multiply. $\endgroup$ – NoChance Mar 18 '16 at 19:37

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