# Math

posted by Gwen

Decide whether each equation has one solution, no solutions, or infinitely many solutions. 3(y-3)=2y-9+y
A. One solution
B. No Solutions
C. Infinitely many solutions
I did 3*y
3*3=9
2y=9+4
3-*=-6
x-27

1. Mia

3(y-3)=2y-9+y

Distribute the 3

3y-9=2y-9+y

Combine the like terms that are on the right side "2y" and "y"

3y-9=3y-9

You can stop right here because the equations are the same on both sides. This tells you that no matter what y is it will be equal on both sides. Which is why the answer is C.

Did I explain clearly?

3(y - 3) = 2y - 9 + y (Distribute on left side, and combine 2y and y on right side.)
3y - 9 = 3y - 9 (Subtract 3y on both sides.)
3y 3y - 3y - 9 = - 9 (Add 9 to both sides.)
3y - 3y = -9 + 9
0 = 0 (Final result.0

So the answer is C because x could be any number and you'll still get 0 as the final result of the equation.

I hope this helps! :)

___3(y - 3) = 2y - 9 + y (Distribute on left side, and combine 2y and y on right side.)
_____3y - 9 = 3y - 9 - 3y(Subtract 3y on both sides.I meant to put - 3y in the right spot in my first post to help you Gwen.)
3y - 3y - 9 = - 9 (Add 9 to both sides.)
____3y - 3y = -9 + 9
__________0 = 0 (Final result.)

So the answer is C because x could be any number and you'll still get 0 as the final result of the equation.

I hope this helps! :)

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