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`