Question

How do you solve by substitution `x + 3y = 5` and `4x + 5y = 13`?

Answer 1

`x+3y=5`
`color(blue)(x=5-3y` ( we substract `3y` on each side )

Now that we have `color(blue)x`, we can substitute it in the second equation :

`4color(blue)x+5y=13`
`4*color(blue)((5-3y))+5y=13`
`20-12y+5y=13`
`20-7y=13`
`20=13+7y` ( we add `7y` on each side )
`20-13=7y`
`7y=7`
`color(red)(y=1` ( we divide by `7` on each side )

Now that we have `color(red)y`, we can find `color(blue)x` :

`color(blue)x=5-3color(red)y`
`color(blue)x=5-3*1`
`color(blue)x=5-3`
`color(blue)(x=2)`