Hi Leo,

I am going to illustrate the method of elimination with a similar problem

\begin{eqnarray*}
5x + 2y & = & 7 \\
15x - 3y & =& 3.
\end{eqnarray*}

I am going to first multiply both sides of the first equation by 3 to get

\begin{eqnarray*}
15x + 6y & = & 21 \\
15x - 3y & =& 3.
\end{eqnarray*}

Notice that the coefficients of $x$ in the two equations are the same. You will see in a moment why this is important.

$15x + 6y$ is $21$ and $15x - 3y$ is $3$ and hence $15x + 6y$ minus $21$ and $15x - 3y$ is $21$ minus $3,$ that is

\[ (15x + 6y) - (15x - 3y) = 21 - 3 \]

or

\[15x + 6y - 15x + 3y = 18.\]

Now you can see why it was important that the coefficients of $x$ in the two equations were the same. The two $x$ terms cancel and we have

\[9x = 18\]

or

\[x = 2.\]

You can now substitute the $x = 2$ into either of the two original equations and solve for $y.$

In my solution I eliminated $x$ but I could have chosen to eliminate $y.$ In the two original equations

\begin{eqnarray*}
5x + 2y & = & 7 \\
15x - 3y & =& 3.
\end{eqnarray*}

if I multiply the first equation by 3 and the second equation by 2 then the coefficients of $y$ become 6 and -6 and hence if I add the resulting equations the y term is eliminated.

I hope this helps,
Penny