To solve the system of equations using the substitution method, we proceed as follows:

Given: $x + y = 2 \quad \text{(Equation 1)}$ $y = x^2 - 10 \quad \text{(Equation 2)}$

Step 1: Substitute Equation 2 into Equation 1

Since $y = x^2 - 10$ from Equation 2, substitute this into Equation 1:

$x + (x^2 - 10) = 2$

Step 2: Simplify the equation

Combine like terms:

$x^2 + x - 10 = 2$

Now, move all terms to one side of the equation:

$x^2 + x - 12 = 0$

Step 3: Solve the quadratic equation

Factor the quadratic equation:

$(x + 4)(x - 3) = 0$

So, the solutions for $x$ are:

$x = -4 \quad \text{or} \quad x = 3$

Step 4: Find the corresponding $y$ values

Now, substitute these $x$ values back into Equation 2 to find the corresponding $y$ values.

1. For $x = -4$: $y = (-4)^2 - 10 = 16 - 10 = 6$ So, one solution is $(-4, 6)$.

2. For $x = 3$: $y = 3^2 - 10 = 9 - 10 = -1$ So, the other solution is $(3, -1)$.

Final Answer

The solution set is: $\{(-4, 6), (3, -1)\}$

Would you like further details or have any questions?

Relative Questions:

1. How do you solve systems of equations using the elimination method?
2. What are the key differences between substitution and elimination methods in solving systems of equations?
3. How do you determine if a system of equations has no solution, one solution, or infinitely many solutions?
4. Can quadratic equations always be factored? If not, what other methods can be used to solve them?
5. What are the advantages of using the substitution method over other methods?

Tip:

When using the substitution method, it's often easier to solve for the variable with the smallest coefficient first to simplify calculations.