Question

question 9 of 25
solve the system of equations:
$y = 2x - 5$
$y = x^2 - 5$

a. $(-1, -4)$ and $(3, 4)$
b. $(-1, -7)$ and $(4, 3)$
c. $(0, 5)$ and $(2, 2)$
d. $(0, -5)$ and $(2, -1)$

Explanation:

Step1: Set the equations equal

Since both equations equal \( y \), set \( 2x - 5 = x^2 - 5 \).
\( 2x - 5 = x^2 - 5 \)

Step2: Simplify the equation

Add 5 to both sides: \( 2x = x^2 \).
Rearrange: \( x^2 - 2x = 0 \).
Factor: \( x(x - 2) = 0 \).

Step3: Solve for \( x \)

Set each factor to zero: \( x = 0 \) or \( x - 2 = 0 \Rightarrow x = 2 \).

Step4: Find corresponding \( y \) values

For \( x = 0 \): Substitute into \( y = 2x - 5 \), \( y = 2(0) - 5 = -5 \). So point is \( (0, -5) \).
For \( x = 2 \): Substitute into \( y = 2x - 5 \), \( y = 2(2) - 5 = -1 \). So point is \( (2, -1) \).

Answer:

D. (0, -5) and (2, -1)