Math Help - linear equation problem (again)

1. linear equation problem (again)

3y+4= y/2+9

how do i do this?! im confused

2. Originally Posted by brainfreeze
3y+4= y/2+9

how do i do this?! im confused
Get the "unknown" (in this case y) by itself on one side of the equation. Here, you might start by getting both "y"s on the left and anything not involving "y" on the right. Since there is a "y/2" added on the left, do the opposite: subtract y/2 from both sides: 3y+ 4- y/2= y/2+ 9- y/2 becomes (3- 1/2)y+ 4= 9 or (6/2- 1/2)y+ 4= (5/2)y+ 4= 9. Now subtract 4 from both sides: (5/2)y+ 4- 4= 9- 4 or (5/2)y= 5. Since y is now multiplied by 5/2, do the opposite: divide both sides by 5/2 (or, same thing, multiply both sides by 2/5): (2/5)(5/2)y= (2/5)(5) or y= 2.

3. thanks, but im stuck again lol,

3/4t= -1 7/8

ive tried to move 7/8 to the other side,
but only to fail.
this is what keeps happening

3/4t= -1 7/8
(3/4÷7/8)t = -1

my answer keeps coming up t=-1 1/6
but the answer is -2 1/6

now i know ive done something wrong
but im not sure what it is lol

4. $\frac{3}{4}t=-\frac{7}{8}$ Is this the problem you want to solve?

5. thats exactly what im trying to solve

6. Originally Posted by SENTINEL4
$\frac{3}{4}t=-\frac{7}{8}$ Is this the problem you want to solve?
Originally Posted by brainfreeze
thats exactly what im trying to solve
@OP: Start by multiplying both sides of the equation by 8:

6t = -7.