I'm not sure how to solve this equation
3(2y−1)−y=2(y+2)−(y−4)

I'm trying to determine whether there is one solution, infinite solutions, or no solutions. The parenthesis are confusing me a little, could somebody give me step-by-step instructions?

Question

I'm not sure how to solve this equation
3(2y−1)−y=2(y+2)−(y−4)

I'm trying to determine whether there is one solution, infinite solutions, or no solutions. The parenthesis are confusing me a little, could somebody give me step-by-step instructions?

avi13 · Answer

3(2y−1)−y=2(y+2)−(y−4)
== 6y-3-y=2y+4-y+4
5y-3=y+8
4y=11
y=11/4

mecharv · Answer

Since no two of the y terms are being multiplied you have only a linear equation with one variable. First let is solve the Right Hand side. 
$$3(2y-1)-y=6y-3-y=5y-3$$

Now the left hand side 

$$2(y+2)-(y-4)=2y+4-y+4=y+8$$

Equating LHS and RHS,

$$y+8=5y-3$$

$$y-5y=-3-8$$
$$-4y=-11$$

$$y=\frac{ -11 }{ -4 }$$
$$y=\frac{ 11 }{ 4 }$$


SO clearly you have only one solution. Cheers :)

mecharv · Answer

@turtwig Still confused?

turtwig · Answer

@mecharv I think I understand now :D Thank you for the thorough response!

mecharv · Answer

Anytime :)