SOLUTION: Hello, I worked this problem and only got half credit. How can I work this as a systems of equations x^2+y^2=25 5=x+y This is what I did x=5-y (y-5)^2+y^2=25 2y-y^2-25=0

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Hello, I worked this problem and only got half credit. How can I work this as a systems of equations x^2+y^2=25 5=x+y This is what I did x=5-y (y-5)^2+y^2=25 2y-y^2-25=0      Log On


   

Question 139467This 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
x^2+y^2=25
5=x+y
This is what I did
x=5-y
(y-5)^2+y^2=25
2y-y^2-25=0 This question is from textbook College Algebra

Found 2 solutions by checkley75, solver91311:
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+y^2=25
5=x+y
x=5-y
(5-y)^2+y^2=25
25-10y+y^2+y^2=25
2y^2-10y=0
2y(y-5)=0
2y=0
y=0 answer.
y-5=0
y=5 answer.
x=5-0
x=5 answer.
x=5-5
y=0 answer.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Think about what you would expect to get before you start working on anything. x%5E2%2By%5E2=25 is a circle centered at the origin with a radius of 5. 5=x%2By can be written as y=-x%2B5 which is a straight line with a slope of -1 and a y-intercept of 5. So we have a straight line and a circle. You have one of three possible outcomes for such a situation. The line intersects the circle in two places, the line is tangent to the circle so it intersects in only one place, or the line doesn't intersect at all.

Given that, your solution method was correct up to the point where you expanded the binomial and collected like terms. Your algebraic manipulation was faulty.

%285-y%29%5E2=25-10y%2By%5E2, so you should have gotten to 25-10y%2By%5E2%2By%5E2-25=0

2y%5E2-10y=0

y%5E2-5y=y%28y-5%29=0

So, y=0 or y+=+5. From here you can calculate the two x-coordinates to complete your points of intersection.

The fact that you substituted y-5 instead of 5-y was an error but not one that would have an impact on the result. y-5=-%285-y%29, and since you were squaring that quantity, the sign wouldn't matter in the final analysis. I can't for the life of me figure out how you got from %28y-5%29%5E2%2By%5E2=25 to 2y-y%5E2-25=0, so my only advice is to go back and thoroughly review the FOIL process for expanding binomials.