Question

3x+4y=23 and 5x+3y=31. Solve by elimination.

Answer 1

Let $3x+4y=23$ be equation 1
let $5x+3y=31$  be equation 2

do $(equ1)*3$ and $(equ2)*4$

now $9x+12y=69$
and $20x+12y=124$
now eliminate the $y$ by doing $(equ2)-(equ1)$
$11x=55$
$x=5$

now substitute $x=5$ back into equation 1 or 2
$3*(5)+4y=23$
$4y=23-15$
$y=2$