1. ## algebra question

I dont know how solve this problem:

Solve 4y+1=2y+8

2. Originally Posted by girliegal
I dont know how solve this problem:

Solve 4y+1=2y+8
our objective here, is to get y on one side of the equal sign by itself. whatever we don't want on one side, we subtract it from both sides of the equation. so let's do that.

$\displaystyle 4y + 1 = 2y + 8$

Let's get the y's on the left side. so i want to get rid of the 2y on the right, so subtract it from both sides.

$\displaystyle \Rightarrow 4y + 1 - 2y = 2y - 2y + 8$

$\displaystyle \Rightarrow 2y + 1 = 8$

Now let's get rid of the 1 on the left side

$\displaystyle \Rightarrow 2y + 1 - 1 = 8 - 1$

$\displaystyle \Rightarrow 2y = 7$

Now divide both sides by 2

$\displaystyle \Rightarrow \frac {2y}{2} = \frac {7}{2}$

$\displaystyle \Rightarrow y = \frac {7}{2}$

3. Thanks for posting.

4. Originally Posted by girliegal
Thanks for posting.
Thanks for clicking the "thanks" button! you understood everything?

5. I didnt understand the end bit.

why did you divide the end bit by two?

6. Originally Posted by girliegal
I didnt understand the end bit.

why did you divide the end bit by two?
remember, the objective was to get y only on one side. at the end i had 2y, i needed to have only y. when i divided by 2, it canceled the 2 that was beside the y, so i was left with y on one side, like i wanted. but since it is an equation, whatever i do on one side, i have to do on the other to keep things equal. so i divided by 2 on the right side as well to get 7/2

7. Originally Posted by Jhevon
remember, the objective was to get y only on one side. at the end i had 2y, i needed to have only y. when i divided by 2, it canceled the 2 that was beside the y, so i was left with y on one side, like i wanted. but since it is an equation, whatever i do on one side, i have to do on the other to keep things equal. so i divided by 2 on the right side as well to get 7/2
k.I understand thx