



 
Hi Leo, I am going to illustrate the method of elimination with a similar problem \begin{eqnarray*} I am going to first multiply both sides of the first equation by 3 to get \begin{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*} 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,
 


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