Question

How do you solve the following system: #y = 3x – 12, x = 4 – 2y #?

Answer 1

`x=4, y=0`

Explanation:

First substitute `x=4-2y` into `y=3x-12`
Then the equation becomes:
`y=3(4-2y)-12`
`y=12-6y-12`
bring all the `y` to the LHS:
`7y=0`
`y=0`
Then substitute `y=0` into `x=4-2y`
`x=4-2(0)`
`x=4`