Question

1. solve the system by substitution.

$x^2 + y^2 = 16$
$y = -x + 4$
options:
a. (8, 2)
b. (0, 4)
c. (6, 12)
d. (4, 0)
e. (4, 2)
f. no real solution

Explanation:

Step1: Substitute \( y = -x + 4 \) into \( x^2 + y^2 = 16 \)

Substitute \( y \) in the circle equation: \( x^2 + (-x + 4)^2 = 16 \)
Expand \( (-x + 4)^2 \): \( x^2 + (x^2 - 8x + 16) = 16 \)
Combine like terms: \( 2x^2 - 8x + 16 = 16 \)
Subtract 16 from both sides: \( 2x^2 - 8x = 0 \)
Factor out \( 2x \): \( 2x(x - 4) = 0 \)

Step2: Solve for \( x \)

Set each factor to zero:

• \( 2x = 0 \Rightarrow x = 0 \)
• \( x - 4 = 0 \Rightarrow x = 4 \)

Step3: Find corresponding \( y \) values

• For \( x = 0 \): \( y = -0 + 4 = 4 \), so the point is \( (0, 4) \)
• For \( x = 4 \): \( y = -4 + 4 = 0 \), so the point is \( (4, 0) \)

Answer:

b. (0, 4), d. (4, 0)