Question

How do you solve `y=sqrt x` and `y=x-6` using substitution?

Answer 1

When `x = 9 , y = 2 " or " 3"`
When `x = 4 , y = 3 " or " -2`

Explanation:

You can solve the equations by equating the y's and then substituting for each y.

`color(red)(y = sqrtx) color(white)(xxxx)and " "color(blue)(y = x-6)`

`color(white)(xxxxxxxxx) color(red)(y) = color(blue)(y)`

Therefore..`color(red)( sqrtx) = color(blue)(x-6)" "larr` only x terms

`(sqrtx)^2 = (x-6)^2 " "larr` square both sides.

`x = x^2 -12x +36" "larr` make the quadratic = 0

`x^2 -13x + 36 = 0" "larr` factorize

Find factors of 36 which ADD to 13. Signs are both negative.

`(x-9)(x-4) = 0`

If `x-9 = 0 rarr x = 9" "` OR If `" "x-4 =0 rarr x = 4`
Now find y.
`color(red)(y = sqrtx) color(white)(xxxx)and " "color(blue)(y = x-6)`

`y = sqrt 9 = 3" "larr` only the principal square root was indicated
`y=sqrt4 = 2`
`y = x-6 rarr y = 9-6 = 3`
`y = x-6 rarr y = 4-6 = -2`