Solve the system of equations $\frac{x}{3} + \frac{y}{2} = \frac{2x}{3} - \frac{y}{6} = 7$ for $x$ and $y$. Here is the problem. The answer is suppose to be x = 12, y = 6
$$\frac{ x }{3} + \frac{y}{2} = \frac{ 2x }{3} - \frac{ y }{6} = 7$$
I've tried this. I'm having trouble figuring out what do on the last step I will describe.
$$6 • \frac{ x }{3} + \frac{y}{2} = \frac{ 2x }{3} - \frac{ y }{6} = 7$$
$$2x + 3y = 4x - y = 42$$
$$2x - 4y = 42$$ 
$$4y = 42 - 2x$$
$$y = \frac{ 42 - 2x}{4}$$
$$2x + 3[ \frac{ 42 - 2x }{4} ] = 7$$
 A: You are really given two equations here:
$$
\left\{\begin{align}
  \frac{x}{3} + \frac{y}{2} &= 7\\[0.3cm]
  \frac{2x}{3} - \frac{y}{6} &= 7
\end{align}
\right.
$$
Multiply the first equation by 6 to get $2x + 3y = 42$.  Multiply the second equation by 6 to get $4x - y = 42$.
$$
\left\{
\begin{align} \displaystyle
  2x + 3y &= 42\\
  4x - y &= 42
\end{align}
\right.
$$
Multiply the second equation by $3$ and add it to the first equation to end up with $14x = 168$, and so $x = 12$.  Then use either equation to get $y = 6$.
A: Your error is linked to how you deal with the two equations (As there are two equal signs you have two equations described in one line). So when you do the following you change from two equations to only one and introduce a mistake when doing so.
Quote:
$$2x + 3y = 4x - y = 42$$
$$2x - 4y = 42$$ 
Firstly you have turned two equations into only one and secondly you have not performed the same operation to all sides of the operation. You have subtracted $2x+3y$ so your working should have looked like:
$$2x + 3y = 4x - y = 42$$
$$0=2x - 4y = 42-2x-3y$$ 
This maintains you have two equations. As a side note it is more generally advisable to work with each equation separately rather than join them together as you have done as you normally want to do different things with each.
Following on from the line above you need to split it into two equations:
$$0=2x-4y \text{  and  } 0=42-2x-3y$$
$$x=2y \text{  and  } 0=42-2x-3y$$
Subbing the first into the second gives:
$$0=42-4y-3y$$
$$7y=42$$
$$y=6$$
Subbing this back into the earlier equation $x=2y$ gives:
$$x=12$$
A: $6·\left[\frac{x}3+\frac{y}2=7\right]$
$18·\left[\frac{2x}3-\frac{y}6=7\right]$
$====================$
then ...
