Question

How do you solve y² = x + 3 and x - 2y = 12?

Answer 1

If `y=5`, `x=22`
If `y=-3`, `x=6`

Explanation:

`y^2=x+3` --- (1)
`x-2y=12` --- (2)

From (2)
`x-2y=12`
`x=12+2y` --- (3)

Sub (3) into (1)
`y^2=12+2y+3`
`y^2=15+2y`
`y^2-2y-15=0`
`(y-5)(y+3)=0`
`y=5` or `y=-3`

If `y=5`,
`x=12+2y`
`x=12+10`
`x=22`

If `y=-3`,
`x=12+2y`
`x=12-6`
`x=6`

Answer 2

`(6,-3),(22,5)`

Explanation:

`y^2=x+3to(1)`

`x-2y=12to(2)`

`"from equation "(2)tox=12+2yto(3)`

`"substitute "x=12+2y" in equation "(1)`

`y^2=12+2y+3`

`y^2-2y-15=0larrcolor(blue)"in standard form"`

`(y-5)(y+3)=0`

`y+3=0rArry=-3`

`y-5=0rArry=5`

`"substitute these values into equation "(3)`

`y=-3tox=12-6=6to(6,-3)`

`y=5tox=12+10=22to(22,5)`

Answer 3

A graphical solution below:

Explanation:

Solving this graphically:

Here's one point of intersection:

graph{(y^2-x-3)(x-2y-12)=0}

And here's the other:

graph{(y^2-x-3)(x-2y-12)=0[10,30,0,10]}