Question

question
find all solutions of the system of equations algebraically. write your solutions as coordinate points.
$y = 2x^2 + 3x - 94$
$3x + y = 14$
answer attempt 1 out of 2
two solutions
and

Explanation:

Step1: Solve for y from the second equation

From \( 3x + y = 14 \), we can express \( y \) as \( y = 14 - 3x \).

Step2: Substitute y into the first equation

Substitute \( y = 14 - 3x \) into \( y = 2x^2 + 3x - 94 \). We get:
\( 14 - 3x = 2x^2 + 3x - 94 \)

Step3: Rearrange into standard quadratic form

Bring all terms to one side:
\( 2x^2 + 3x + 3x - 94 - 14 = 0 \)
\( 2x^2 + 6x - 108 = 0 \)
Divide through by 2:
\( x^2 + 3x - 54 = 0 \)

Step4: Factor the quadratic equation

We need two numbers that multiply to -54 and add to 3. These numbers are 9 and -6. So,
\( (x + 9)(x - 6) = 0 \)

Step5: Solve for x

Set each factor equal to zero:
\( x + 9 = 0 \) or \( x - 6 = 0 \)
So, \( x = -9 \) or \( x = 6 \).

Step6: Find the corresponding y - values

For \( x = -9 \):
Substitute into \( y = 14 - 3x \), \( y = 14 - 3(-9)=14 + 27 = 41 \).
For \( x = 6 \):
Substitute into \( y = 14 - 3x \), \( y = 14 - 3(6)=14 - 18=-4 \).

Answer:

\((-9, 41)\) and \((6, -4)\)