Answer
$(10, 68)$ and $(-3, 16)$
Work Step by Step
We will use substitution to solve this system of equations. We substitute one of the expressions given for the $y$ term, which would mean that we are going to set the two equations equal to one another to solve for $x$ first:
$x^2 - 3x - 2 = 4x + 28$
We want to move all terms to the left side of the equation:
$x^2 - 3x - 4x - 2 - 28 = 0$
Combine like terms:
$x^2 - 7x - 30 = 0$
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 that sum up to $b$. In this exercise, $ac = -30$ and $b = -7$. The factors $-10$ and $3$ will work:
$(x - 10)(x + 3) = 0$
Set each factor equal to $0$.
First factor:
$x - 10 = 0$
Add $10$ to each side of the equation:
$x = 10$
Second factor:
$x + 3 = 0$
Subtract $3$ from each side of the equation:
$x = -3$
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:
$y = 4x + 28$
Substitute the solution $10$ for $x$:
$y = 4(10) + 28$
Multiply next:
$y = 40 + 28$
Add to solve:
$y = 68$
Let's solve for $y$ using the other solution, $x = -3$:
$y = 4(-3) + 28$
Multiply first:
$y = -12 + 28$
Add to solve:
$y = 16$
The solutions are $(10, 68)$ and $(-3, 16)$.
