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 Question from leo, a student: please explain how can i solve this problem 3x-6y=-38 6x-9y=44 using elimination and simultaneous method thank you :)

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

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