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
\(3*(5)+4y=23\)
\(4y=23-15\)
\(y=2\)
