Question

question
find all solutions of the system of equations algebraically.
write your solutions as coordinate points.

$y = 2x^2 - 5x - 136$
$3x + y = 8$

answer
two solutions
and

Explanation:

Step1: Solve for y from the linear equation

From \(3x + y = 8\), we can express \(y\) as \(y = 8 - 3x\).

Step2: Substitute y into the quadratic equation

Substitute \(y = 8 - 3x\) into \(y = 2x^2 - 5x - 136\). So we get:
\(8 - 3x=2x^2 - 5x - 136\)

Step3: Rearrange the equation to standard quadratic form

Bring all terms to one side:
\(2x^2 - 5x + 3x- 136 - 8 = 0\)
Simplify to:
\(2x^2 - 2x - 144 = 0\)
Divide the entire equation by 2:
\(x^2 - x - 72 = 0\)

Step4: Factor the quadratic equation

We need two numbers that multiply to \(- 72\) and add up to \(-1\). The numbers are \(-9\) and \(8\). So the factored form is:
\((x - 9)(x + 8)=0\)

Step5: Solve for x

Set each factor equal to zero:
\(x - 9 = 0\) gives \(x = 9\)
\(x + 8 = 0\) gives \(x=-8\)

Step6: Find the corresponding y - values

For \(x = 9\), substitute into \(y = 8 - 3x\):
\(y=8-3\times9=8 - 27=-19\)
For \(x=-8\), substitute into \(y = 8 - 3x\):
\(y = 8-3\times(-8)=8 + 24 = 32\)

Answer:

The solutions as coordinate points are \((9,-19)\) and \((-8,32)\)