Question

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solve the following system of equations algebraically:

$y = x^2 + 7x - 19$

$y = 3x + 2$

answer
attempt 1 out of 2
two solutions
and

Explanation:

Step1: Set equations equal

Since both equal $y$, equate them:
$x^2 + 7x - 19 = 3x + 2$

Step2: Rearrange to standard quadratic

Move all terms to left side:
$x^2 + 7x - 19 - 3x - 2 = 0$
$x^2 + 4x - 21 = 0$

Step3: Factor the quadratic

Find two factors of -21 that sum to 4:
$(x + 7)(x - 3) = 0$

Step4: Solve for $x$

Set each factor equal to 0:
$x + 7 = 0 \implies x = -7$
$x - 3 = 0 \implies x = 3$

Step5: Find corresponding $y$ values

Substitute $x=-7$ into $y=3x+2$:
$y = 3(-7) + 2 = -19$
Substitute $x=3$ into $y=3x+2$:
$y = 3(3) + 2 = 11$

Answer:

$(-7, -19)$ and $(3, 11)$