To find the solution(s) of the system of equations represented by [tex]\( y = x^2 + 2x - 3 \)[/tex] and [tex]\( y = x - 1 \)[/tex], we need to determine the points where these two equations intersect each other. These points of intersection represent the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously.
Let's solve this step-by-step:
1. Set the equations equal to each other: Since both equations are equal to [tex]\( y \)[/tex], we set them equal to each other to find the [tex]\( x \)[/tex]-values where they intersect.
[tex]\[
x^2 + 2x - 3 = x - 1
\][/tex]
2. Rearrange the equation: Move all terms to one side of the equation to set it to zero.
[tex]\[
x^2 + 2x - 3 - x + 1 = 0
\][/tex]
Simplify this to:
[tex]\[
x^2 + x - 2 = 0
\][/tex]
3. Factor the quadratic equation: Factor the quadratic expression to find the values of [tex]\( x \)[/tex].
[tex]\[
x^2 + x - 2 = (x + 2)(x - 1) = 0
\][/tex]
4. Solve for [tex]\( x \)[/tex]: Set each factor equal to zero and solve for [tex]\( x \)[/tex].
[tex]\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\][/tex]
[tex]\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\][/tex]
5. Find corresponding [tex]\( y \)[/tex]-values: Substitute these values of [tex]\( x \)[/tex] back into either original equation (we'll use [tex]\( y = x - 1 \)[/tex] for simplicity) to find the corresponding [tex]\( y \)[/tex]-values.
For [tex]\( x = -2 \)[/tex]:
[tex]\[
y = -2 - 1 = -3
\][/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[
y = 1 - 1 = 0
\][/tex]
6. Write the solution points: The points of intersection, which are solutions to the system, are:
[tex]\[
(-2, -3) \quad \text{and} \quad (1, 0)
\][/tex]
Therefore, the solutions to the system of equations [tex]\( y = x^2 + 2x - 3 \)[/tex] and [tex]\( y = x - 1 \)[/tex] are [tex]\( (-2, -3) \)[/tex] and [tex]\( (1, 0) \)[/tex].
