SOLUTION: Hello, I worked this problem and only got half credit. How can I work this as a systems of equations y=x^2-4 x-y=-2 This is what I did x=y-2 y=(y-2)^2-4 2y-4-4=0

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Question 139463This question is from textbook College Algebra
: Hello, I worked this problem and only got half credit.
How can I work this as a systems of equations
y=x^2-4
x-y=-2
This is what I did
x=y-2
y=(y-2)^2-4
2y-4-4=0
This question is from textbook College Algebra

Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
Start with the given system
y=x%5E2-4
x-y=-2


x=y-2 Solve for x in the second equation


y=%28y-2%29%5E2-4 Plug in x=y-2


y=y%5E2-4y%2B4-4 Foil


0=y%5E2-4y%2B4-4-y Subtract y from both sides


0=y%5E2-5y Combine like terms


0=y%28y-5%29 Factor the right side




Now set each factor equal to zero:
y=0 or y-5=0

y=0 or y=5 Now solve for y in each case


So our y-values are

y=0 or y=5


Let's find x when y=0

x=y-2 Start with the second equation


x=0-2 Plug in y=0


x=-2 Subtract


So when x=-2 then y=0



Let's find x when y=5

x=y-2 Start with the second equation


x=5-2 Plug in y=5


x=3 Subtract


So when x=3 then y=5




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Answer:

So the solutions are

(-2,0) or (3,5)