Solve Quadratic System
x^2+y^2=2
x^2-6y=10
Please SHOW WORK I'm having trouble
To solve the given quadratic system:
1. Rearrange the second equation to isolate one of the variables in terms of the other:
x^2 - 6y = 10
x^2 = 6y + 10
x = ±√(6y + 10)
2. Substitute the value of x from step 1 into the first equation:
(±√(6y + 10))^2 + y^2 = 2
6y + 10 + y^2 = 2
3. Rearrange the above equation to form a quadratic equation in one variable:
y^2 + 6y + 8 = 0
4. Solve the quadratic equation in step 3 by factoring or using the quadratic formula. In this case, the equation factors to:
(y + 2)(y + 4) = 0
Setting each factor equal to zero:
y + 2 = 0 --> y = -2
y + 4 = 0 --> y = -4
5. Substitute the values of y from step 4 into either equation to find the corresponding value(s) of x:
For y = -2:
x = ±√(6(-2) + 10)
x = ±√(-2) --> No real solutions
For y = -4:
x = ±√(6(-4) + 10)
x = ±√(-14) --> No real solutions
Therefore, the given quadratic system has no real solutions.