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Mathematics 23 Online
OpenStudy (anonymous):

How to solve 5y(2y-3)+(2y-3)

OpenStudy (anonymous):

What's common to both terms? an example: 2y(5+1)= 5*2y+2y

OpenStudy (anonymous):

What do you mean.? the 2y-3 is

OpenStudy (anonymous):

yep, so take it out as a common factor, and that should give you something like 2y(3y+3). (these are different numbers). I'm guessing the equation is equal to zero so for a multiplication to be 0 either of the terms multiplying must be 0. first in this case would be y=0 and y=-1

OpenStudy (anonymous):

If I took it out as a common factor wouldn't it just cancel out.? or were you basically saying 2y(3y+3) and 3(2y+2)?

OpenStudy (anonymous):

Sorry I was using a different equation. Try factoring out the common thing in your equation and then try to get the multiplication to be zero, like in my example

OpenStudy (anonymous):

Okay I'm lost

OpenStudy (anonymous):

5y(2y-3)+(2y-3)=0 take the common factor and write it out

OpenStudy (anonymous):

2y(5+1)= 5*2y+2y

OpenStudy (anonymous):

same idea as here

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