Answer
The solution is $(1, -2)$.
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:
$2x^2 - 3x - 1 = x - 3$
We want to move all terms to the left side of the equation.
$2x^2 - 3x - x - 1 + 3 = 0$
Combine like terms:
$2x^2 - 4x + 2 = 0$
Factor out any common terms:
$2(x^2 - 2x + 1) = 0$
Divide each side by $2$:
$x^2 - 2x + 1 = 0$
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, $ac = 1$ and $b = -2$. The factors $-1$ and $-1$ will work:
$(x - 1)(x - 1) = 0$
Use the Zero-Product Property by equating each unique factor to $0$, then solve each equation.
$x - 1 = 0$
Add $1$ to each side of the equation:
$x = 1$
Now that we have the possible value for $x$, we can plug them into one of the original equations to find the corresponding $y$ value. Let's use the second equation:
$y = x - 3$
Substitute the solution $1$ for $x$:
$y = 1 - 3$
Subtract to solve:
$y = -2$
The solution is $(1, -2)$.
