How do you solve #x^2 + y^2 = 32# and #y - x = 0#? Algebra Systems of Equations and Inequalities Systems Using Substitution 1 Answer anton Sep 17, 2016 #x_1=4: y_1=4; x_2=-4; y_2=-4# Explanation: #y-x=0 rArr y=x# #x^2+x^2=32; 2x^2=32: x^2=16; x_1=4; x_2=-4# Answer link Related questions How do you solve systems of equations using the substitution method? How do you check your solutions to a systems of equations using the substitution method? When is the substitution method easier to use? How do you know if a solution is "no solution" or "infinite" when using the substitution method? How do you solve #y=-6x-3# and #y=3# using the substitution method? How do you solve #12y-3x=-1# and #x-4y=1# using the substitution method? Which method do you use to solve the system of equations #y=1/4x-14# and #y=19/8x+7#? What are the 2 numbers if the sum is 70 and they differ by 11? How do you solve #x+y=5# and #3x+y=15# using the substitution method? What is the point of intersection of the lines #x+2y=4# and #-x-3y=-7#? See all questions in Systems Using Substitution Impact of this question 5719 views around the world You can reuse this answer Creative Commons License