Question

solve the system of equations.
$y = x^2 + 40x - 6$
$y = 28x + 39$
write the coordinates in exact form. simplify all fractions and radicals.
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Explanation:

Step1: Set the equations equal

Since both equal \( y \), set \( x^2 + 40x - 6 = 28x + 39 \).

Step2: Simplify the equation

Subtract \( 28x + 39 \) from both sides: \( x^2 + 12x - 45 = 0 \).

Step3: Solve the quadratic equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for \( ax^2 + bx + c = 0 \). Here, \( a = 1 \), \( b = 12 \), \( c = -45 \).
\( x = \frac{-12 \pm \sqrt{12^2 - 4(1)(-45)}}{2(1)} = \frac{-12 \pm \sqrt{144 + 180}}{2} = \frac{-12 \pm \sqrt{324}}{2} = \frac{-12 \pm 18}{2} \).

Step4: Find x-values

First solution: \( x = \frac{-12 + 18}{2} = \frac{6}{2} = 3 \).
Second solution: \( x = \frac{-12 - 18}{2} = \frac{-30}{2} = -15 \).

Step5: Find y-values

For \( x = 3 \), \( y = 28(3) + 39 = 84 + 39 = 123 \).
For \( x = -15 \), \( y = 28(-15) + 39 = -420 + 39 = -381 \).

Answer:

\((3, 123)\)
\((-15, -381)\)