Answer
The solutions are $(3, 7)$ and $(-2, 2)$.
Work Step by Step
We will use substitution to solve this system of equations. We substitute one of the expressions for $y$, which would mean that we are going to set the two equations equal to one another to solve for $x$ first:
$-x^2 + 2x + 10 = x + 4$
We want to move all terms to the left side of the equation.
$-x^2 + 2x - x + 10 - 4 = 0$
Combine like terms:
$-x^2 + x + 6 = 0$
Divide both sides by $-1$:
$x^2 - x - 6 = 0$
Factor the quadratic equation. The quadratic equation takes the form $ax^2 + bx + c = 0$. We need to find factors whose product is $ac$ but sum up to $b$. In this exercise, $ac = -6$ and $b = -1$. The factors $-3$ and $2$ will work:
$(x - 3)(x + 2) = 0$
Use the Zero-Product Property by equating each factor to $0$, then solve each equation.
First factor:
$x - 3 = 0$
$x = 3$
Second factor:
$x + 2 = 0$
$x = -2$
Now that we have the two possible values for $x$, we can plug them into one of the original equations to find the corresponding $y$ values. Let's use the second equation:
$y = x + 4$
Substitute the solution $3$ for $x$:
$y = 3 + 4$
Multiply next:
$y = 7$
Let's solve for $y$ using the other solution, $x = -2$:
$y = -2 + 4$
Add to solve:
$y = 2$
The solutions are $(3, 7)$ and $(-2, 2)$.
