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Find the solution set of:
y=1/3(x-3)^2 - 3
(x-3)^2 + (y+2)^2 = 1
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You are given the system of two equations
y = , (1)
= 1. (2)
Multiply the equation (1) by 3 to rid off the denominator. You will get
3y = (1') instead of (1).
From (1') express = 3y + 9. Based on it, replace in (2) by 3y + 9. You will get
= 1. (3)
Simplify (3) and solve for "y":
= ,
= .
Factor left side
(y+3)*(y+4) = 0.
The solutions are y = -3 and/or y = -4.
1. For y = -3 we have from (1') = = = 1 - 1 = 0. Hence, x = 3.
2. For y = -4 we have from (1') = = = 1 - 4 = -3, and there are no real solutions.
Answer. x = 3, y = -3.
For solving systems of non-linear equations see the lessons
- Solving systems of algebraic equations of degree 2 and degree 1
- Solving systems of algebraic equations of degree 2
- Solving systems of non-linear algebraic equations with symmetric functions of unknowns
in this site.