# How do you solve 4(2y-1)=4(y+3) for y?

Mar 25, 2015

Less thought, more mechanical: first remove the parentheses on both sides by distributing the multiplication. Then collect all the terms involving $y$ on one side (say the left) and the terms without $y$ on the right. Finally, divide both sides by the coefficient of $y$. (the number in front of $y$).

It looks like this:

$4 \left(2 y - 1\right) = 4 \left(y + 3\right)$

$8 y - 4 = 4 y + 12$

$8 y - 4 y = 12 + 4$ (subtract $4 y$ from both sides and add $4$ to both sides)

$4 y = 16$

$\frac{4 y}{4} = \frac{16}{4}$

$y = 4$

More thought, looking at the equation:

Instead of distributing the $4$'s on both sides, you could start by dividing both sides by $4$ (multiplying by $\frac{1}{4}$. This keeps the numbers smaller. It looks like this:
$4 \left(2 y - 1\right) = 4 \left(y + 3\right)$

$\frac{1}{4} \left[4 \left(2 y - 1\right)\right] = \frac{1}{4} \left[4 \left(y + 3\right]\right)$

#1/4*4 = [1/4 *4] (y+3)

$2 y - 1 = y + 3$

$2 y - y = 3 + 1$ (subtract $y$ and add $1$ on both sides)

$y = 4$

Mar 25, 2015

First multiply $4$ to get rid of the brackets:
$8 y - 4 = 4 y + 12$
Now, isolate the $y$s on one side and numbers on the other of the $=$ sign (remembering to change sign in crossing the $=$ sign);
$8 y - 4 y = + 4 + 12$
Notice the change of sign of $4 y$ and $- 4$;
now we can add and subtract:
$4 y = 16$
The $4$ is multiplying the $x$ and can go to the right dividing to get:
$x = \frac{16}{4} = 4$

hope it helps!