Solve. http://static.k12.com/bank_packages/files/media/mathml_f55cd31001ee25e0467b0eb517efa3c1e62c61db_1.gif y =
@Godlovesme
@freckles
solve for y
first step: See if you multiply both sides by 3 so we can get rid of that ugly fraction
let me know what you get after performing just this step
sorry for you wait lag how do i do that sorry to be bother
I want you to try to multiply both sides on 3 and tell me what you think you have as a result of doing that
so you mean 3x1?
3(y-1)=4y-2
ok
Second step: use distributive property where possible
that is multiply 3(y-1)
3y-3=4y-2, right?
great \[y-1=\frac{4y-2}{3} \\ \text{ Multiply both sides by 3 } \\ 3(y-1)=4y-2 \\ \text{ distribute on the left hand side } \\ 3y-3=4y-2\]
so now we want to get our terms with y on one side and the terms without y on the other side
There is a few ways to start out doing this.
Like I might decide to subtract 3y on both sides then add 2 on both sides... You might decide you are going to subtract 4y on both sides then add 3 on both sides. Either way here is fine.
so 4y+3y?
hmmm... how did you get that?
did you try to subtract 3y on both sides ?
\[-3=4y-3y-2\]
4y-3y=1y
yes now add 2 on both sides
\[y-1=\frac{4y-2}{3} \\ \text{ Multiply both sides by 3 } \\ 3(y-1)=4y-2 \\ \text{ Distribute on the left hand side } \\ 3y-3=4y-2 \\ \text{ Subtract } 3y \text{ on both sides } \\ -3=4y-3y-2 \\ \text{ add 2 on both sides } \\ -3+2=4y-3y \]
-3-2=-5
you added 2 on both sides
oh you said add on give sec
so you should have -3+2
ok
i got -1
ok sounds great \[ y-1=\frac{4y-2}{3} \\ \text{ Multiply both sides by 3 } \\ 3(y-1)=4y-2 \\ \text{ Distribute on the left hand side } \\ 3y-3=4y-2 \\ \text{ Subtract } 3y \text{ on both sides } \\ -3=4y-3y-2 \\ \text{ add 2 on both sides } \\ -3+2=4y-3y \\ -1=y \]
so the answer is -1?
yes
Thanks:)
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