# What are the solution(s) to the following system of equations #y = -x^2# and y = x?

x = 0 , x = - 1

Since we are given 2 values that y equates to we can equate the right sides.

By signing up, you agree to our Terms of Service and Privacy Policy

The solution(s) to the system of equations y = -x^2 and y = x are the points where the two equations intersect.

Substitute y = x into the first equation: x = -x^2

Rearrange the equation into standard quadratic form: x^2 + x = 0

Factor out x: x(x + 1) = 0

So, either x = 0 or x + 1 = 0, which gives x = 0 or x = -1.

When x = 0, y = 0. When x = -1, y = -1.

Therefore, the solution(s) to the system of equations are (0, 0) and (-1, -1).

By signing up, you agree to our Terms of Service and Privacy Policy

- How do you solve the following system: # -6x + 5y = 4 , x-10y=10 #?
- How do you solve by substitution #y = 2x + 4# and #y = 3x #?
- If A,B, and C are nxn matrices, does AB=AC mean B = C?
- On the situation where taking the numbers 123456 how many numbers can u form using 3 digits with no numbers repeated is that a permutation or combination?
- How do you solve the system of equations #y= - 2x + 6# and #2y + x = - 5#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7