Question

find all solutions of the system of equations algebraically. write your solutions as coordinate points.
$y = -3x^2 + 30x - 41$
$3x + y = 13$
answer
two solutions
and
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Explanation:

Step1: Solve the linear equation for y

From \(3x + y = 13\), we get \(y = 13 - 3x\).

Step2: Substitute y into the quadratic equation

Substitute \(y = 13 - 3x\) into \(y = -3x^2 + 30x - 41\):
\(13 - 3x = -3x^2 + 30x - 41\)

Step3: Rearrange into standard quadratic form

\(3x^2 - 33x + 54 = 0\) (divide all terms by 3)
\(x^2 - 11x + 18 = 0\)

Step4: Factor the quadratic equation

\((x - 2)(x - 9) = 0\)

Step5: Solve for x

Set each factor equal to zero:
\(x - 2 = 0\) gives \(x = 2\)
\(x - 9 = 0\) gives \(x = 9\)

Step6: Find corresponding y values

For \(x = 2\), \(y = 13 - 3(2) = 7\)
For \(x = 9\), \(y = 13 - 3(9) = -14\)

Answer:

\((2, 7)\) and \((9, -14)\)