SOLUTION: Solve by graphing: x^2+y^2=25 and x-2y=-5 I graphed my circle and line perfectly but when it came to using either addition or substitution methods to find the solution set (wh

Algebra ->  Algebra  -> Graphs -> SOLUTION: Solve by graphing: x^2+y^2=25 and x-2y=-5 I graphed my circle and line perfectly but when it came to using either addition or substitution methods to find the solution set (wh      Log On

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Question 116265: Solve by graphing: x^2+y^2=25 and x-2y=-5
I graphed my circle and line perfectly
but when it came to using either addition or substitution methods
to find the solution set (where the line crosses the circle at two points)
I had problems:(
I can see by looking at my graph that one solution is (-5,0) and the other
is approximately (3.5, 4.5) but I have to use one of the above methods
to show how I came up with my solutions and show my work.
This is what I've done so far:
Using substitution method
x^2+y^2=25 and x-2y=-5 (changed to x=2y-5) and substituted in to 1st equation
(2y-5)^2+y^2=25
4y^2-20y+25+y^2=25
(combine like terms)
5y^2-20y+25=25
(subtract 25 from each side)
5y^2-20y=0
This is where I loose it, because in class all of our problems worked out nice and neat
and all I had to do was factor my quadratic equation to come up with one solution and
then plug that into one of my original problems to get my second.

No one in study group tonight quite got this one either and this is due this Thursday!
Found 2 solutions by stanbon, bucky:
Answer by stanbon(57940) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+y^2=25
x-2y=-5
---------
x = 2y-5
Substitute to get:
(2y-5)^2 + y^2 = 25
4y^2-20y+25 + y^2 = 25
5y^2-20y=0
5y(y-4) = 0
y = 0 or y = 4
------------------
Substitute y = 0 into x-2y=-5 to get x=-5
Substitute y = 4 into x-2y=-5 to get x-8 = -5; x = 3
-------------------
Solution: (-5,0) and (3,4)
=====================
Cheers,
Stan H.

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Good Job ...
.
x^2+y^2=25 and x-2y=-5 (changed to x=2y-5) and substituted in to 1st equation
(2y-5)^2+y^2=25 <=== OK
.
4y^2-20y+25+y^2=25 <=== OK
.
(combine like terms)
5y^2-20y+25=25 <=== OK
.
(subtract 25 from each side)
5y^2-20y=0 <====
.
Now just factor 5y out of the two terms on the left side to get:
.
5y*(y - 4) = 0
.
This equation will be true if either of the two factors on the left side equals zero because
a multiplication by zero on the left side makes the left side zero and, therefore,
equal to the zero on the right side.
.
So there are two possible answers for y (because the line crosses the circle at two points).
.
Set the factors equal to zero ...
.
First:
.
5y = 0 ... divide both sides by 5 to get y = 0
.
Next:
.
y - 4 = 0 ... add 4 to both sides ...
.
y = 4
.
Plug these two values for y (0 and 4) into the equation x - 2y = -5. [Of the two equations this will
probably be the easier one to work with.]
.
When y = 0 the equation x - 2y = -5 becomes x = -5. So (-5, 0) is one of the intersection
points.
.
When y = 4 the equation x - 2y = -5 becomes x - (2*4) = -5 which simplifies to x - 8 = -5
.
Add 8 to both sides and you get x = +3. So the point (3, 4) is the second point at which the
line crosses the circle.
.
You did a very good job on the hardest part of the problem ... you just needed a little hint
at the very last part. Keep up the good work ... good luck to you and your study group.
.