# How do you solve the system #x^2-2x+2y+2=0# and #-x^2+2x-y+3=0#?

##### 1 Answer

Jun 3, 2018

#### Explanation:

#x^2-2x+2y+2=0to(1)#

#-x^2+2x-y+3=0to(2)#

#"add equations "(1)" and "(2)" term by term"#

#(x^2-x^2)+(-2x+2x)+(2y-y)+5=0#

#y+5=0rArry=-5#

#"substitute "y=-5" into either "(1)" or "(2)" and"#

#"solve for x"#

#(1)tox^2-2x-10+2=0#

#x^2-2x-8=0larrcolor(blue)"in standard form"#

#(x-4)(x+2)=0#

#x-4=0rArrx=4#

#x+2=0rArrx=-2#

#"solutions are "(-2,-5)" or "(4,-5)#

graph{(x^2-2x+2y+2)(x^2-2x+y-3)=0 [-10, 10, -5, 5]}