SOLUTION: Hello. I need to know how you multiply this expression: (x-2)/(x+3) * (4x+12)/(6). Now, I've gotten to the step where the problem is simplified to (x-2)/(1) * (4)/(6). I don't real

Question 592321: Hello. I need to know how you multiply this expression: (x-2)/(x+3) * (4x+12)/(6). Now, I've gotten to the step where the problem is simplified to (x-2)/(1) * (4)/(6). I don't really know how to go about multiplying that expression, so could you please help? I have a big math test tomorrow and need to understand these concepts before it's too late. Thank you!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
multiply this expression: (x-2)/(x+3) * (4x+12)/(6).
:
*
Factoring will help here before we multiply
*
We can cancel (x+3) and then we have
(x - 2) *
Reduce the fraction, then multiply
(x - 2) * = or
:
Generally you multiply the numerators, and multiply the denominators. then reduce the fraction
But factoring and canceling can greatly simplify the operation

RELATED QUESTIONS
Hello, I need to solve this equation. If there is no solution I have to say so. 5/4x + (answered by stanbon)

I am trying to solve the quadratic equation using the method from India. I need... (answered by stanbon)

In this expression I know that the first step is to find the common denominator. And... (answered by stanbon)

I am really frustrated with this problem, can you help me? 5/x+7 + 4/x-3 =... (answered by Boreal)

(2x+3)/4 + (x-5) / 3 = 3/2 I have done some of this but i have gotten stuck after this (answered by Cromlix)

This is actually a Calculus 1 problem, but I am having problems with the algebra portion. (answered by solver91311)

Can you please show me step by step how to get this answer for the problem -1/2-5/6 x =... (answered by josgarithmetic,Edwin McCravy)

3x+4=12x+5 This is what I"ve tried but don't know if I'm right. Plus I need to know... (answered by josgarithmetic,MathTherapy)

Hello, I'm taking Calculus online and have a homework problem I'm having trouble with. (answered by Alan3354)