Question

1
select the correct answer from each drop - down menu.
$y = x^{2}+2x - 1$
$y - 3x = 5$
the pair of points representing the solution set of this system of equations is and .
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Explanation:

Step1: Express y from the linear equation

From \( y - 3x = 5 \), we get \( y = 3x + 5 \).

Step2: Substitute y into the quadratic equation

Substitute \( y = 3x + 5 \) into \( y = x^2 + 2x - 1 \):
\( 3x + 5 = x^2 + 2x - 1 \)

Step3: Rearrange into standard quadratic form

\( x^2 + 2x - 1 - 3x - 5 = 0 \)
\( x^2 - x - 6 = 0 \)

Step4: Factor the quadratic equation

Factor \( x^2 - x - 6 \): \( (x - 3)(x + 2) = 0 \)

Step5: Solve for x

Set each factor to zero:
\( x - 3 = 0 \) gives \( x = 3 \)
\( x + 2 = 0 \) gives \( x = -2 \)

Step6: Find corresponding y values

For \( x = 3 \): \( y = 3(3) + 5 = 14 \), so the point is \( (3, 14) \)
For \( x = -2 \): \( y = 3(-2) + 5 = -1 \), so the point is \( (-2, -1) \)

Answer:

\((-2, -1)\) and \((3, 14)\)