If x+y = 1, and x^2 + y^2 = -1, what is x^7 + y^13?
1) 0
2) 1
3) -2
4) 2
5) -1

Solve for x and y :

x + y = 1   →   y = 1 - x

x ² + y ² = -1

x ² + (1 - x)² = -1

x ² + (1 - 2x + x ²) = -1

2x ² - 2x + 1 = -1

2x ² - 2x + 2 = 0

x ² - x + 1 = 0

x ² - x + 1/4 = -3/4

(x - 1/2)² = -3/4

x - 1/2 = ±√(-3/4)

x - 1/2 = ±√3/2 i

x = 1/2 ± √3/2 i   →   x = exp(± iπ/3)

y = 1 - (1/2 ± √3/2 i )   →   y = -1/2 ± √3/2 i   →   y = exp(± 2iπ/3)

Then

x ⁷ + y ¹³ = exp(± 7iπ/3) + exp(± 26iπ/3)

… = exp(± iπ/3) + exp(± 2iπ/3)

since 7π/3 is equivalent to π/3, and 26π/3 is equivalent to 2π/3 (both modulo 2π).

In either case, we get

x ⁷ + y ¹³ = x + y = 1

so the answer is (2) 1.

42

step-by-step explanation:

c.

step-by-step explanation:

substitution for (0,1)

y=x^2+4x+1

1=0^2+4(x)0+1

1=0+4(x)0+1

1=4(x)0+1

1=0+1

1=1

substitution for (-3,-2)

y=x+1

-2=-3+1

-2=-2