Question

find all solutions of the system of equations algebraically. write your solutions as coordinate points.
$y = x^2 + 3x - 41$
$16 = 3x - y$
answer
two solutions
and
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Explanation:

Step1: Solve the second equation for y

From \(16 = 3x - y\), we can rearrange it to get \(y = 3x - 16\).

Step2: Substitute y into the first equation

Substitute \(y = 3x - 16\) into \(y = x^2 + 3x - 41\). So we have \(3x - 16 = x^2 + 3x - 41\).

Step3: Simplify the equation

Subtract \(3x\) from both sides: \(-16 = x^2 - 41\). Then add 41 to both sides: \(x^2 = 25\).

Step4: Solve for x

Take the square root of both sides: \(x = \pm 5\).

Step5: Find the corresponding y values

• When \(x = 5\), substitute into \(y = 3x - 16\): \(y = 3(5) - 16 = 15 - 16 = -1\).
• When \(x = -5\), substitute into \(y = 3x - 16\): \(y = 3(-5) - 16 = -15 - 16 = -31\).

Answer:

\((5, -1)\) and \((-5, -31)\)