SOLUTION: Find the solution set of: y=1/3(x-3)^2

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Question 1045648: Find the solution set of:
y=1/3(x-3)^2 - 3
(x-3)^2 + (y+2)^2 = 1
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850)   (Show Source):
You can put this solution on YOUR website!

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if you set , then ........substitute it in

it will be true if
and it will be true if
so, real solution to the system above is: ,
or point (,)

Answer by ikleyn(52817)   (Show Source):
You can put this solution on YOUR website!
.
Find the solution set of:
y=1/3(x-3)^2 - 3
(x-3)^2 + (y+2)^2 = 1
~~~~~~~~~~~~~~~~~~~~~~~

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.

SOLUTION: Find the solution set of: y=1/3(x-3)^2 - 3 (x-3)^2 + (y+2)^2 = 1