Question

solve the system of equations.
$y = 23x + 46$
$y = x^2 + 41x + 6$
write the coordinates in exact form. simplify all fractions and radicals.
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Explanation:

Step1: Set equations equal

Since both equal \( y \), set \( 23x + 46 = x^2 + 41x + 6 \).

Step2: Rearrange into quadratic

Subtract \( 23x + 46 \) from both sides: \( x^2 + 18x - 40 = 0 \).

Step3: Solve quadratic (quadratic formula)

For \( ax^2 + bx + c = 0 \), \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 18 \), \( c = -40 \).
Discriminant: \( 18^2 - 4(1)(-40) = 324 + 160 = 484 \).
\( \sqrt{484} = 22 \).
\( x = \frac{-18 \pm 22}{2} \).
Two solutions:
\( x_1 = \frac{-18 + 22}{2} = \frac{4}{2} = 2 \).
\( x_2 = \frac{-18 - 22}{2} = \frac{-40}{2} = -20 \).

Step4: Find corresponding \( y \)

For \( x = 2 \): \( y = 23(2) + 46 = 46 + 46 = 92 \).
For \( x = -20 \): \( y = 23(-20) + 46 = -460 + 46 = -414 \).

Answer:

\((2, 92)\)
\((-20, -414)\)