Answer
$\left(-\frac{4}{3}, 4\right)$ and $(-8, -16)$
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:
$-\frac{3}{4}x^2 - 4x = 3x + 8$
We want to move all terms to the left side of the equation:
$-\frac{3}{4}x^2 - 4x - 3x - 8 = 0$
Combine like terms:
$-\frac{3}{4}x^2 - 7x - 8 = 0$
Multiply both sides of the equation by $-1$ so that the $x^2$ term is positive:
$\frac{3}{4}x^2 + 7x + 8 = 0$
Multiply both sides of the equation by $4$ to convert all fractions into whole numbers:
$3x^2 + 28x + 32 = 0$
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, $ac = 96$ and $b = 28$. The factors $24$ and $4$ will work. Let's split the middle term using these factors:
$3x^2 + 24x + 4x + 32 = 0$
Group the first two terms and the last two terms:
$(3x^2 + 24x) + (4x + 32) = 0$
Factor out any common terms:
$3x(x + 8) + 4(x + 8) = 0$
Group the factors:
$(3x + 4)(x + 8) = 0$
Set each factor equal to $0$.
First factor:
$3x + 4 = 0$
Subtract $4$ from each side of the equation:
$3x = -4$
Divide each side by $3$:
$x = -\frac{4}{3}$
Second factor:
$x + 8 = 0$
Subtract $8$ from each side of the equation:
$x = -8$
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:
$y = 3x + 8$
Substitute the solution $-\frac{4}{3}$ for $x$:
$y = 3(-\frac{4}{3}) + 8$
Multiply:
$y = -\frac{12}{3} + 8$
Simplify the fraction:
$y = -4 + 8$
Add to solve:
$y = 4$
Let's solve for $y$ using the other solution, $x = -8$:
$y = 3(-8) + 8$
Multiply:
$y = -24 + 8$
$y = -16$
The solutions are $\left(-\frac{4}{3}, 4\right)$ and $(-8, -16)$.
