To solve this system of equations algebraically, let's continue from the given steps:
### Step 1: Set the Equations Equal
We start by setting the two equations equal to each other:
[tex]\[ y = x^2 - x - 3 \][/tex]
[tex]\[ y = -3x + 5 \][/tex]
[tex]\[ x^2 - x - 3 = -3x + 5 \][/tex]
### Step 2: Move All Terms to One Side
Next, we need to move all the terms to one side of the equation to set it to 0:
[tex]\[ x^2 - x - 3 + 3x - 5 = 0 \][/tex]
[tex]\[ x^2 + 2x - 8 = 0 \][/tex]
### Step 3: Factor the Quadratic Equation
We factor the quadratic equation:
[tex]\[ (x - 2)(x + 4) = 0 \][/tex]
### Step 4: Solve for [tex]\( x \)[/tex]
We set each factor to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x - 2 = 0 \implies x = 2 \][/tex]
[tex]\[ x + 4 = 0 \implies x = -4 \][/tex]
### Step 5: Find Corresponding [tex]\( y \)[/tex] Values
We now substitute the x-values back into one of the original equations to find the corresponding y-values. Let's use [tex]\( y = x^2 - x - 3 \)[/tex]:
For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 2^2 - 2 - 3 \][/tex]
[tex]\[ y = 4 - 2 - 3 \][/tex]
[tex]\[ y = -1 \][/tex]
For [tex]\( x = -4 \)[/tex]:
[tex]\[ y = (-4)^2 - (-4) - 3 \][/tex]
[tex]\[ y = 16 + 4 - 3 \][/tex]
[tex]\[ y = 17 \][/tex]
### Final Solutions
So, the solution set to the system of equations is:
[tex]\[ (2, -1) \][/tex]
[tex]\[ (-4, 17) \][/tex]
Thus, the complete solution is [tex]\((2, -1)\)[/tex] and [tex]\((-4, 17)\)[/tex].
