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Find the solution(s) of the system of equations: x2 + y2 = 8 y = x – 4 options: (–2,–6) (2,–2) and (–2,–6) (2,–2) No solutions

User Rodorgas
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4.9k points

2 Answers

6 votes

Answer:

(2,–2)

Explanation:

User Deandob
by
5.0k points
5 votes

Answer: x=2 y=-2

(2,-2) one solution

Step-by-step explanation:

Solve by substitution


\begin{bmatrix}x^2+y^2=8\\ y=x-4\end{bmatrix}


\mathrm{Subsititute\:}y=x-4


\begin{bmatrix}x^2+\left(x-4\right)^2=8\end{bmatrix}


2x^2-8x+16=8


\mathrm{Isolate}\:x\:\mathrm{for}\:2x^2-8x+16=8:\quad x=2


\mathrm{For\:}y=x-4


\mathrm{Subsititute\:}x=2


y=2-4
2-4=-2


y=-2


The\:solutions\:to\:the\:system\:of\:equations\:are


x=2,\:y=-2

User Zimmerbimmer
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