To solve the system of equations given by:
1. [tex]\( y = x^2 + 2x - 1 \)[/tex]
2. [tex]\( y - 3x = 5 \)[/tex]
we follow these steps:
### Step-by-Step Solution
Step 1: Express y in terms of x using the second equation
From the second equation [tex]\( y - 3x = 5 \)[/tex], we can solve for [tex]\( y \)[/tex]:
[tex]\[ y = 3x + 5 \][/tex]
Step 2: Substitute [tex]\( y \)[/tex] from the second equation into the first equation
Now we substitute [tex]\( y \)[/tex] in the first equation with [tex]\( 3x + 5 \)[/tex]:
[tex]\[ 3x + 5 = x^2 + 2x - 1 \][/tex]
Step 3: Rearrange the equation to form a standard quadratic equation
Rearrange the equation to have all terms on one side:
[tex]\[ x^2 + 2x - 1 - 3x - 5 = 0 \][/tex]
[tex]\[ x^2 - x - 6 = 0 \][/tex]
Step 4: Factorize the quadratic equation
We need to factorize [tex]\( x^2 - x - 6 \)[/tex]:
[tex]\[ x^2 - x - 6 = (x - 3)(x + 2) = 0 \][/tex]
Step 5: Solve for x
Solve the factored equation for [tex]\( x \)[/tex]:
[tex]\[ x - 3 = 0 \quad \text{or} \quad x + 2 = 0 \][/tex]
[tex]\[ x = 3 \quad \text{or} \quad x = -2 \][/tex]
Step 6: Find the corresponding values of y
Using the values of [tex]\( x \)[/tex] found, substitute back into the expression [tex]\( y = 3x + 5 \)[/tex] to find the corresponding [tex]\( y \)[/tex]-values:
For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 3(3) + 5 \][/tex]
[tex]\[ y = 9 + 5 \][/tex]
[tex]\[ y = 14 \][/tex]
For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = 3(-2) + 5 \][/tex]
[tex]\[ y = -6 + 5 \][/tex]
[tex]\[ y = -1 \][/tex]
### Conclusion
So, the pair of points that represents the solution set of the system of equations is:
[tex]\[ (3, 14) \quad \text{and} \quad (-2, -1) \][/tex]
Thus, the pair of points representing the solution set of this system of equations is:
[tex]\[ \boxed{(3, 14) \quad \text{and} \quad (-2, -1)} \][/tex]
