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

1 Answer
Sep 29, 2016

Answer:

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#