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

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Answer 1 · 2016-09-29T18:25:14

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

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
$y=sqrt4 = 2$
$y = x-6 rarr y = 9-6 = 3$
$y = x-6 rarr y = 4-6 = -2$

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