How do you solve the system of equations: #y = x^2 + 9# and #y = 6x#?

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Answer 1 · 2017-10-31T15:56:29

See a solution process below:

Because both equations are stated in terms of #y# we can equation the right side of each equation and solve for #x#:

#x^2 + 9 = 6x#

#x^2 - color(red)(6x) + 9 = 6x - color(red)(6x)#

#x^2 - 6x + 9 = 0#

#(x - 3)(x - 3) = 0#

Because both terms on the left side of the equation are the same we can equate one of them to #0# to solve for #x#:

#x - 3 = 0#

#x - 3 + color(red)(3) = 0 + color(red)(3)#

#x - 0 = 3#

#x = 3#

Now, we can substitute #3# for #x# in either of the equations to calculate #y#:

#y = x^2 + 9# becomes:

#y = 3^2 + 9 = 9 + 9 = 18#

Or

#y = 6x# becomes:

#y = 6 xx 3 = 18#

The Solution Is: #x = 3# and #y = 18# or #(3, 18)#