Arthur D. answered • 05/24/16
Tutor
4.9
(293)
Forty Year Educator: Classroom, Summer School, Substitute, Tutor
x^2-y^2=3
2x+y^2=5
y^2=-2x+5
x^2-(-2x+5)=3
x^2+2x-5=3
x^2+2x-5-3=0
x^2+2x-8=0
(x+4)(x-2)=0
x+4=0
x=-4
x-2=0
x=2
now substitute -4 for x and find y
then substitute 2 for x and find y
each x value will give you two y values
Alex H.
actually I think you did it wrong even if it looks correct. I have the answer sheet and it says the answer is
(2,1) (2,-1) (-4,√13), (-4,-√13)
Report
05/24/16
Arthur D.
tutor
I did it right. I left it for you to finish.
x=-4
(-4)^2-y^2=3
16-y^2=3
16-3=y^2
13=y^2
y=√13 or √-13
(-4,√13) and (-4,√-13)
2x+y^2=5
2(-4)+y^2=5
-8+y^2=5
y^2=13
y=±√13
x=2
2x+y^2=5
4+y^2=5
y^2=1
y=±1
(2,1), and (2,-1)
You see. The problem is done correctly.
Report
05/25/16
Alex H.
05/24/16