Question

solve each system

1. $y = x^2 + 3x - 23$

$y = 25 - 5x$

1. $y = x^2 + 2x - 2$

$y = x + 10$

1. $y = x^2 + 4$

$y = 4x$

1. $y = x^2 + x + 3$

$y = -x$

Explanation:

First System: \( y = x^2 + 3x - 23 \) and \( y = 25 - 5x \)

Step1: Set equations equal

Since both equal \( y \), set \( x^2 + 3x - 23 = 25 - 5x \).

Step2: Rearrange to quadratic

\( x^2 + 8x - 48 = 0 \).

Step3: Factor the quadratic

\( (x + 12)(x - 4) = 0 \).

Step4: Solve for \( x \)

\( x = -12 \) or \( x = 4 \).

Step5: Find \( y \) for \( x = -12 \)

Substitute \( x = -12 \) into \( y = 25 - 5x \): \( y = 25 - 5(-12) = 85 \).

Step6: Find \( y \) for \( x = 4 \)

Substitute \( x = 4 \) into \( y = 25 - 5x \): \( y = 25 - 5(4) = 5 \).

Step1: Set equations equal

\( x^2 + 2x - 2 = x + 10 \).

Step2: Rearrange to quadratic

\( x^2 + x - 12 = 0 \).

Step3: Factor the quadratic

\( (x + 4)(x - 3) = 0 \).

Step4: Solve for \( x \)

\( x = -4 \) or \( x = 3 \).

Step5: Find \( y \) for \( x = -4 \)

Substitute \( x = -4 \) into \( y = x + 10 \): \( y = 6 \).

Step6: Find \( y \) for \( x = 3 \)

Substitute \( x = 3 \) into \( y = x + 10 \): \( y = 13 \).

Step1: Set equations equal

\( x^2 + 4 = 4x \).

Step2: Rearrange to quadratic

\( x^2 - 4x + 4 = 0 \).

Step3: Factor the quadratic

\( (x - 2)^2 = 0 \).

Step4: Solve for \( x \)

\( x = 2 \) (double root).

Step5: Find \( y \) for \( x = 2 \)

Substitute \( x = 2 \) into \( y = 4x \): \( y = 8 \).

Answer:

Solutions: \( (-12, 85) \), \( (4, 5) \)

Second System: \( y = x^2 + 2x - 2 \) and \( y = x + 10 \)