How do you solve #x² + y² = 20# and #x + y = 6# using substitution?

1 Answer
Sep 5, 2017

Answer:

See a solution process below:

Explanation:

Step 1) Solve the second equation for #x#:

#x + y = 6#

#x + y - color(red)(y) = 6 - color(red)(y)#

#x + 0 = 6 - y#

#x = 6 - y#

Step 2) Substitute #(6 - y)# for #x# in the first equation and solve for #y#:

#x^2 + y^2 = 20# becomes:

#(6 - y)^2 + y^2 = 20#

#36 - 12y + y^2 + y^2 = 20#

#1y^2 + 1y^2 - 12y + 36 = 20#

#(1 + 1)y^2 - 12y + 36 = 20#

#2y^2 - 12y + 36 = 20#

#2y^2 - 12y + 36 - color(red)(20) = 20 - color(red)(20)#

#2y^2 - 12y + 16 = 0#

#(2y - 8)(y - 2) = 0#

Solution 1 for #y#:

#2y - 8 = 0#

#2y - 8 + color(red)(8) = 0 + color(red)(8)#

#2y - 0 = 8#

#2y = 8#

#(2y)/color(red)(2) = 8/color(red)(2)#

#(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = 4#

#y = 4#

Solution 2 for #y#:

#y - 2 = 0#

#y - 2 + color(red)(2) = 0 + color(red)(2)#

#y - 0 = 2#

#y = 2#

Step 3) Substitute the values for #y# into the solution to the second equation at the end of Step 1 and calculate #x#:

#x = 6 - y# and #x = 6 - y# becomes:

#x = 6 - 4# and #x = 6 - 2#

#x = 2# and #x = 4#

The Solutions Are:

#x = 2# and #y = 4# or #(2, 4)#

AND

#x = 4# and #y = 2# or #(4, 2)#