You can put this solution on YOUR website! 2x + y = 10
x^2+y^2= 25
:
We are going to use the 1st equation for substituting in the 2nd equation:
2x + y = 10
y = (10-2x)
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Substitute (10-2x) for y in the 2nd equation:
x^2 + (10-2x)^2 = 25
:
FOIL (10-2x)(10-2x); then we have:
x^2 + 100 - 40x + 4x^2 = 25
:
x^2 + 4x^2 - 40x + 100 - 25 = 0
:
5x^2 - 40x + 75 = 0
:
Simplify, divide equation by 5
x^2 - 8x + 15 = 0
:
Factor:
(x-5)(x-3) = 0
:
x = 5; and x = 3
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Find y when x = 5:
y = 10 - 2x
y = 10 -2(5)
y = 0
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Find y when x = 3
y = 10 - 2(3)
Y = 4
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We have two sets of solutions:
x = 5; y = 0
and
x = 3; y = 4
:
You can confirm both of these solutions in both equations
