SOLUTION: x² + 2y² - 4x + 6y - 5 = 0, x + y + 5 = 0 Solve by method of substitution. Thank you.

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Question 1080928: x² + 2y² - 4x + 6y - 5 = 0,
x + y + 5 = 0
Solve by method of substitution. Thank you.
Found 2 solutions by Fombitz, ikleyn:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
x=-y-5=-%28y%2B5%29
x%5E2=%28-%28y%2B5%29%29%5E2=y%5E2%2B10y%2B25
Substituting,
y%5E2%2B10y%2B25%2B2y%5E2-4%28-%28y%2B5%29%29%2B6y-5=0
y%5E2%2B10y%2B25%2B2y%5E2%2B4y%2B20%2B6y-5=0
3y%5E2%2B20y%2B40=0
This equation has no real solutions.
So there is no solution to the system.
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Verified by the graph since there is no intersection.
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Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
See the lesson
    - Solving the system of algebraic equations of degree 2 and degree 1,
in this site. You will find there a lot of similar solved problems.


Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic "Systems of equations that are not linear".