Question

solve each system of equations using substitution.

1. $y = x^{2}+x - 60$

$y = 2x - 4$

1. $y = x^{2}-3x + 7$

$y = 4x - 3$

1. $y = x^{2}-2x - 5$

$y = x - 5$

Explanation:

Step1: Substitute \(y\) in equation 7

Since \(y = 2x - 4\) and \(y=x^{2}+x - 60\), we set \(2x-4=x^{2}+x - 60\).
\[x^{2}+x - 60-(2x - 4)=0\]
\[x^{2}+x - 60 - 2x + 4=0\]
\[x^{2}-x - 56=0\]

Step2: Factor the quadratic equation

Factor \(x^{2}-x - 56\) as \((x - 8)(x+7)=0\).

Step3: Solve for \(x\)

Set each factor equal to zero:
If \(x - 8=0\), then \(x = 8\); if \(x+7=0\), then \(x=-7\).

Step4: Find \(y\) - values

When \(x = 8\), \(y=2\times8 - 4=16 - 4 = 12\).
When \(x=-7\), \(y=2\times(-7)-4=-14 - 4=-18\).

Step5: Substitute \(y\) in equation 8

Since \(y = 4x - 3\) and \(y=x^{2}-3x + 7\), we set \(4x-3=x^{2}-3x + 7\).
\[x^{2}-3x + 7-(4x - 3)=0\]
\[x^{2}-3x + 7 - 4x+3=0\]
\[x^{2}-7x + 10=0\]

Step6: Factor the quadratic equation

Factor \(x^{2}-7x + 10\) as \((x - 2)(x - 5)=0\).

Step7: Solve for \(x\)

Set each factor equal to zero:
If \(x - 2=0\), then \(x = 2\); if \(x - 5=0\), then \(x = 5\).

Step8: Find \(y\) - values

When \(x = 2\), \(y=4\times2-3=8 - 3=5\).
When \(x = 5\), \(y=4\times5-3=20 - 3 = 17\).

Step9: Substitute \(y\) in equation 9

Since \(y=x - 5\) and \(y=x^{2}-2x - 5\), we set \(x - 5=x^{2}-2x - 5\).
\[x^{2}-2x - 5-(x - 5)=0\]
\[x^{2}-2x - 5 - x+5=0\]
\[x^{2}-3x=0\]

Step10: Factor the quadratic equation

Factor \(x^{2}-3x\) as \(x(x - 3)=0\).

Step11: Solve for \(x\)

Set each factor equal to zero:
If \(x=0\), then \(y=0 - 5=-5\); if \(x - 3=0\), then \(x = 3\) and \(y=3 - 5=-2\).

Answer:

For equation 7: \((8,12)\) and \((-7,-18)\)
For equation 8: \((2,5)\) and \((5,17)\)
For equation 9: \((0,-5)\) and \((3,-2)\)