# How do you solve 4x+7y=6 and 6x+5y=20 using elimination?

Nov 9, 2014

$\left\{\begin{matrix}4 x + 7 y = 6 \\ 6 x + 5 y = 20\end{matrix}\right.$

by multiplying the first equation by $3$ and the second equation by $2$,

$\implies \left\{\begin{matrix}12 x + 21 y = 18 \\ 12 x + 10 y = 40\end{matrix}\right.$

by subtracting the second equation from the first equation,

$\implies 11 y = - 22$

by dividing by $11$,

$\implies y = - 2$

By plugging $y = - 2$ into the first equation of the original system,

$4 x + 7 \left(- 2\right) = 6 \implies 4 x - 14 = 6$

by adding $14$,

$\implies 4 x = 20$

by dividing by $4$,

$\implies x = 5$

Hence, the solution is $\left(x , y\right) = \left(5 , - 2\right)$.

I hope that this was helpful.