[tex]\bf \begin{cases}
\quad3x+2y=-9\\
-10x+5y=-5
\end{cases}[/tex]
so hmm the idea behind the elimination, is to make the value atop or below, of either variable, the same but with a negative sign
so hmmm say... we'll eliminate "y"
so... let's see, we have 2y and 5y, both positive
what do we multiply 2y to get a negative -5y, that way, -5y +5y = 0 and poof it goes
well let's see, let's try "k" [tex]\bf 2yk=-5y\implies k=-\cfrac{5y}{2y}\implies k=-\cfrac{5}{2}[/tex]
so.. if we multiply 2y by -5/2, we'll end up with -5y
well, let's multiply that first equation then by -5/2
[tex]\bf \begin{array}{rllll}
3x+2y=-9&\qquad\times -\frac{5}{2}\implies &-\frac{15}{2}x-5y=\frac{45}{2}\\\\
-10x+5y=-5&\implies &-10x+5y=-5\\
&&---------\\
&&\frac{-35}{2}x+0\quad=\frac{35}{2}
\end{array}\\\\
-----------------------------\\\\
\cfrac{-35}{2}x=\cfrac{-35}{2}\implies \cfrac{-35x}{2}=\cfrac{-35}{2}\implies x=\cfrac{2}{-35}\cdot \cfrac{-35}{2}
\\\\\\
\boxed{x=1}[/tex]
so.. now we know x = 1.... so let's plug that in say hmmm 2nd equation
[tex]\bf -10(1)+5y=-5\implies -10+5y=-5\implies 5y=5\implies\boxed{ y=1}[/tex]
