Question

How do you solve the following system: #4x-y=6, 5x+2y=20 #?

Answer 1

`x = 32/13 and y = 50/13`

Explanation:

`" "4xcolor(blue)(-y)=6" ".........A`
`" "5xcolor(blue)(+2y)=20" ".....B`

One of the easiest methods to eliminate one of the variables is to make them into additive inverses. (Their sum is `0`)

Multiplying equation `A` by `2` will achieve this,:

`Axx2:color(white)(xx)8xcolor(blue)(-2y)=12" ".........C`
`color(white)(xxxxxxx)5xcolor(blue)(+2y)=20" ".....B`

`C+B:color(white)(xx)13x" "=32" ".........C`
`C+B:color(white)(x..x)x" "=32/13" "`

If `x=32/13`, substitute into one of the original equations to find `y`

`5(32/13)+2y=20" ".....B`

`160/13+2y=20`

`" "2y = 20-160/13`

`" "2y = 100/13`

`" "y=50/13`

Check in `A`

`4(32/13) -50/13`
`=78/13`
`=6`