To locate the solutions of the given system of equations:
[tex]\[
\begin{array}{l}
y = x^2 + x - 2 \\
y = -x + 1
\end{array}
\][/tex]
We need to find the points where the two curves intersect. An intersection point [tex]\((x, y)\)[/tex] is a solution to both equations simultaneously. Let's solve this step-by-step.
### Step 1: Set Equations Equal to Each Other
Since both equations are equal to [tex]\( y \)[/tex], we can set the right-hand sides of the equations equal to each other:
[tex]\[ x^2 + x - 2 = -x + 1 \][/tex]
### Step 2: Rearrange and Combine Like Terms
Move all the terms to one side of the equation to set the equation to 0:
[tex]\[ x^2 + x + x - 2 - 1 = 0 \][/tex]
[tex]\[ x^2 + 2x - 3 = 0 \][/tex]
### Step 3: Solve the Quadratic Equation
We will solve the quadratic equation by factoring:
[tex]\[ x^2 + 2x - 3 = 0 \][/tex]
To factor, we look for two numbers that multiply to [tex]\(-3\)[/tex] and add to [tex]\(2\)[/tex].
These numbers are [tex]\(3\)[/tex] and [tex]\(-1\)[/tex]:
[tex]\[ (x + 3)(x - 1) = 0 \][/tex]
Setting each factor equal to zero gives us:
[tex]\[ x + 3 = 0 \][/tex]
[tex]\[ x = -3 \][/tex]
and
[tex]\[ x - 1 = 0 \][/tex]
[tex]\[ x = 1 \][/tex]
### Step 4: Find Corresponding y-values
We now substitute these [tex]\( x \)[/tex]-values back into either original equation to find the corresponding [tex]\( y \)[/tex]-values. Let's use [tex]\( y = -x + 1 \)[/tex]:
For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = -(-3) + 1 \][/tex]
[tex]\[ y = 3 + 1 = 4 \][/tex]
So, the first solution is [tex]\((-3, 4)\)[/tex].
For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = -1 + 1 \][/tex]
[tex]\[ y = 0 \][/tex]
So, the second solution is [tex]\((1, 0)\)[/tex].
### Conclusion
The solutions to the system of equations [tex]\(\begin{array}{l} y = x^2 + x - 2 \\ y = -x + 1 \end{array}\)[/tex] are:
[tex]\[
\boxed{(-3, 4) \text{ and } (1, 0)}
\][/tex]
Thus, the correct choice from the list of options is:
[tex]\[
(-3, 4) \text{ and } (1, 0)
\][/tex]
