# How do I use elimination to find the solution of the system of equations 4x+3y=7 and 3x+5y=8?

May 19, 2015

$4 x + 3 y = 7$ ..........equation $\left(1\right)$
$3 x + 5 y = 8$ ..........equation $\left(2\right)$

multiplying equation 1 by 3 and equation 2 by 4 :

$\left(4 x + 3 y = 7\right) \times 3$
$12 x + 9 y = 21$........equation $\left(3\right)$

and:

$\left(3 x + 5 y = 8\right) \times 4$
$12 x + 20 y = 32$.......equation $\left(4\right)$

on subtracting equation $4$ from $3$,
$12 x$ gets cancelled

$\cancel{12} x + 9 y = 21$
$- \cancel{12} x - 20 y = - 32$

$- 11 y = - 11$
$y = 1$

substituting this value of $y$ in equation 1:

$4 x + 3 \left(1\right) = 7$
$4 x = 7 - 3$
$4 x = 4$
$x = 1$

the solutions for the system of equations are :
$\textcolor{b l u e}{x} = 1$
$\textcolor{b l u e}{y} = 1$