Solve algebraically and check
y = x^2 - 3
x + y = -1
y appears in both equations to the first degree. You can eliminate y by equating the value of y from each equation.
Solve the resulting quadratic equation for x.
Back-substitute the values of x into the equations to get the corresponding values of y.
Substitute x^2 - 3 for y in the second equation.
x + x^2 - 3 = -1
x^2 + x - 2 = 0
(x + 2)(x - 1) = 0
Find the two possible values for x and insert to find y. Insert both sets of values into second equation to check.
To solve the system of equations algebraically, we can substitute the value of y from the first equation into the second equation. Let's start:
1. Start with the equation "y = x^2 - 3".
2. Replace the 'y' in the second equation with "x^2 - 3" from the first equation.
So we have: x + (x^2 - 3) = -1.
3. Simplify the equation: x + x^2 - 3 = -1.
4. Move all terms to one side to get a quadratic equation: x^2 + x - 2 = 0.
5. We can solve this quadratic equation by factoring or using the quadratic formula. Let's use factoring in this case:
Factor the quadratic equation: (x - 1)(x + 2) = 0.
6. Set each factor equal to zero and solve for x:
x - 1 = 0 --> x = 1
x + 2 = 0 --> x = -2.
7. We have two possible values for x: x = 1 or x = -2.
8. To find the corresponding y-values, substitute the x-values into any of the original equations. Let's use the first equation.
For x = 1: y = (1)^2 - 3 = -2.
For x = -2: y = (-2)^2 - 3 = 1.
9. Therefore, the solutions to the system of equations are (1, -2) and (-2, 1).
To check if these solutions are correct, substitute the x and y values into both original equations and see if they satisfy the equations.
Let's substitute the values of (1, -2) into the equations:
For the first equation: y = x^2 - 3
Substituting x = 1 and y = -2, we get: -2 = (1)^2 - 3
Simplifying, -2 = 1 - 3
-2 = -2 ✔
For the second equation: x + y = -1
Substituting x = 1 and y = -2, we get: 1 + (-2) = -1
Simplifying, -1 = -1 ✔
Since both equations are satisfied, the solution (1, -2) is correct.
Now let's substitute the values of (-2, 1) into the equations:
For the first equation: y = x^2 - 3
Substituting x = -2 and y = 1, we get: 1 = (-2)^2 - 3
Simplifying, 1 = 4 - 3
1 = 1 ✔
For the second equation: x + y = -1
Substituting x = -2 and y = 1, we get: -2 + 1 = -1
Simplifying, -1 = -1 ✔
Both equations are satisfied, so the solution (-2, 1) is also correct.
Therefore, the solutions to the system of equations are (1, -2) and (-2, 1).