Answer
$(-3, -4)$ and $(1, 8)$
Work Step by Step
First, we need to get the $y$ term by itself for both equations:
$y = 3x + 5$
$y = x^2 + 5x + 2$
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:
$x^2 + 5x + 2 = 3x + 5$
We want to move all terms to the left side of the equation:
$x^2 + 5x - 3x + 2 - 5 = 0$
Combine like terms:
$x^2 + 2x - 3 = 0$
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, $ac = -3$ and $b = 2$. The factors $3$ and $-1$ will work:
$(x + 3)(x - 1) = 0$
Set each factor equal to $0$.
First factor:
$x + 3 = 0$
Subtract $3$ to each side of the equation:
$x = -3$
Second factor:
$x - 1 = 0$
Add $1$ to each side of the equation:
$x = 1$
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 first equation:
$y = 3x + 5$
Substitute the solution $-3$ for $x$:
$y = 3(-3) + 5$
Multiply next:
$y = -9 + 5$
Add to solve:
$y = -4$
Let's solve for $y$ using the other solution $x = 1$:
$y = 3(1) + 5$
Multiply first:
$y = 3 + 5$
Add to solve:
$y = 8$
The solutions are $(-3, -4)$ and $(1, 8)$.
