SOLUTION: Given the equation of the circle {{{ x^2 +y^2=25 }}} and the equation of the line {{{ 2x+y=10}}}... Find the solutions algebraically. I don't get what we're supposed to do with tho
Question 1007316: Given the equation of the circle and the equation of the line ... Find the solutions algebraically. I don't get what we're supposed to do with those equations, add them, multiply, or what? Found 2 solutions by ankor@dixie-net.com, Edwin McCravy:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Given the equation of the circle x^2 + y^2 = 25 and
the equation of the line 2x + y = 10...
Find the solutions algebraically
:
One way is write the 2nd equation as
y = -2x + 10
Substitute for y in the 1st equation
x^2 + (-2x+10)^2 = 25
FOIL (-2x+10(-2x+10)
x^2 + (4x^2 - 20x -20x + 100) = 25
Combine like terms
x^2 + 4x^2 - 40x + 100 - 25 = 0
5x^2 - 40x + 75 = 0
Simplify, divide by 5
x^2 - 8x + 15 = 0
Factors to
(x-5)(x-3) = 0
Two solutions
x = 5
x = 3
:
Using the equation y = -2x+10, find y using both x solution
y = -2(5) + 10
y = 0
and
y = -2(3) + 10
y = -6 + 10
y = 4
:
Solutions:
x=5; y=0
x=3; y=4
:
Check in the 1st equation using the last pair, (the first pair is obvious)
3^2 + 4^2 = 25
9 + 16 = 25
Instead of doing your problem for you, I'll
do one EXACTLY like it in every detail, step
by step. You can then do yours EXACTLY the
same way, using this one as a guide.
Given the equation of an ellipse and the
equation of a line ...Find the solutions
algebraically.
Solve the second equation for y
Substitute for y in the first equation:
Factor:
5x-14=0; x-2=0
5x=14; x=2
x=14/5;
Substitute each in
Substitute x=14/5
So one point is (x,y) = (14/5,1/5)
Substitute x=2
So the other point is (x,y) = (2,-1)
Now do yours exactly the same way
step-by-step.
Edwin
