Solve the system
x^2+(y-3)^2-9=0
x+y-1=0
Substitute 1-x for y in the first equation.
x^2 + (x+2)^2 = 9
4x + 4 = 9
Take it from there
To solve the system of equations:
Step 1: Substitute 1 - x for y in the first equation.
- x^2 + (1 - x - 3)^2 - 9 = 0
- x^2 + (-x - 2)^2 - 9 = 0
- x^2 + x^2 + 4x + 4 - 9 = 0
- 2x^2 + 4x - 5 = 0
Step 2: Simplify the equation obtained in Step 1.
- 2x^2 + 4x - 5 = 0
Step 3: Solve the quadratic equation by factoring or using the quadratic formula.
The quadratic equation factors as:
- (2x - 1)(x + 5) = 0
Setting each factor equal to zero gives two possible solutions:
- 2x - 1 = 0 or x + 5 = 0
Solving for x in each equation:
- 2x - 1 = 0
2x = 1
x = 1/2
- x + 5 = 0
x = -5
Therefore, the two possible values of x are x = 1/2 and x = -5.
Step 4: Substitute the values of x obtained in Step 3 into the second equation to find the corresponding values of y.
For x = 1/2:
y = 1 - x
= 1 - 1/2
= 1/2
For x = -5:
y = 1 - x
= 1 - (-5)
= 6
Therefore, the corresponding values of y are y = 1/2 and y = 6.
Step 5: Check the solution by substituting the values of x and y obtained in Step 4 into the first equation of the original system.
For x = 1/2 and y = 1/2:
x^2 + (y - 3)^2 - 9 = 0
(1/2)^2 + (1/2 - 3)^2 - 9 = 0
1/4 + (-5/2)^2 - 9 = 0
1/4 + 25/4 - 9 = 0
26/4 - 36/4 = 0
-10/4 = 0
-5/2 = 0 [False]
For x = -5 and y = 6:
x^2 + (y - 3)^2 - 9 = 0
(-5)^2 + (6 - 3)^2 - 9 = 0
25 + 3^2 - 9 = 0
25 + 9 - 9 = 0
25 = 0 [False]
Neither solution satisfies the original system of equations. Therefore, the system has no solution.