Answer
The solutionsare $(1, 2)$ and $(2, 3)$.
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:
$x^2 - 2x + 3 = x + 1$
We want to move all terms to the left side of the equation.
$x^2 - 2x - x + 3 - 1 = 0$
Combine like terms:
$x^2 - 3x + 2 = 0$
Factor the quadratic equation. We need to find factors whose product is $ac$, which is $2$ in this case, but sum up to $b$, which is $-3$ in this case. The factors $-2$ and $-1$ will work:
$(x - 2)(x - 1) = 0$
Set each factor equal to $0$.
First factor:
$x - 2 = 0$
Add $2$ to each side of the equation:
$x = 2$
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 second equation:
$y = x + 1$
Substitute the solution $1$ for $x$:
$y = 1 + 1$
Multiply next:
$y = 2$
Let's solve for $y$ using the other solution $x = 2$:
$y = 2 + 1$
Add to solve:
$y = 3$
The solutions are $(1, 2)$ and $(2, 3)$.
