Solving for $x$ given $y = 2x - 9$ and $y = 5$. When does the 5 come into play? I am doing some problems. It states "Isolate the left side for each of the following equations. Then solve for x, assuming the value of y is 5 in a all cases." I noticed that I can get two different answers, depending on when I decide to plug in the value of y. For example, this is what I get:
$$ y = 2x - 9 $$
$$ 9 + y = 2x -9 +9 $$
$$ 9 + y = 2x \frac{1}{2} $$
$$ \frac{1}{2} * 9 + y = x $$
$$ x = \frac{1}{2} * 9 + 5 $$
$$ x = \frac{9}{2} + 5 $$
$$ x = \frac{19}{2} $$
But, if 5 is plugged into y in the beginning, I've seen it done like this:
$$ 5 = 2x - 9 $$
$$ 5 + 9 = (2x - 9) + 9 $$
$$ 14 = 2x -9 + 9 $$
$$ 14 * \frac{1}{2} = 2x * \frac{1}{2} $$
$$ x = 7 $$
Which one is correct?

The first is wrong because you forgot to multiply $y$ by $\tfrac12$. When you got into
$$
9 + y = 2x
$$
By the law of equations, you should do as follows:
$$
(9 + y) = 2x \\
\frac12(9+y) = \frac12 2x \\
\frac12(9+y) = x
$$
The you plug the value of $y$
$$
\frac12(9 + 5) = x \\
\frac12 14 = x \\
x = 7
$$
