Answer: x=2 y=-2
(2,-2) one solution
Step-by-step explanation:
Solve by substitution
[tex]\begin{bmatrix}x^2+y^2=8\\ y=x-4\end{bmatrix}[/tex]
[tex]\mathrm{Subsititute\:}y=x-4[/tex]
[tex]\begin{bmatrix}x^2+\left(x-4\right)^2=8\end{bmatrix}[/tex]
[tex]2x^2-8x+16=8[/tex]
[tex]\mathrm{Isolate}\:x\:\mathrm{for}\:2x^2-8x+16=8:\quad x=2[/tex]
[tex]\mathrm{For\:}y=x-4[/tex]
[tex]\mathrm{Subsititute\:}x=2[/tex]
[tex]y=2-4[/tex] [tex]2-4=-2[/tex]
[tex]y=-2[/tex]
[tex]The\:solutions\:to\:the\:system\:of\:equations\:are[/tex]
[tex]x=2,\:y=-2[/tex]