1. Find all values of $x$ such that the fraction x/(x - 5) = the fraction 4/(x - 4) 2. The roots of the quadratic equation $z^2 + bz + c = 0$ are $-7 + 2i$ and $-7 - 2i$. What is $b+c$?
3. Find one pair $(x, y)$ of real numbers such that $x + y = 4$ and $x^3 + y^3 = 100.$

1. x/(x - 5) = 4/(x - 4)

x (x - 4) / ((x - 5) (x - 4)) = 4 (x - 5) / ((x - 5) (x - 4))

x (x - 4) = 4 (x - 5)

x² - 4x = 4x - 20

x² - 8x = -20

Solving for x by completing the square gives

x² - 8x + 16 = -4

(x - 4)² = -4

x - 4 = ± 2i

x = 4 + 2i   or   x = 4 - 2i

2. Since -7 + 2i and -7 - 2i are roots of the given quadratic, we have

z² + bz + c = (z - (-7 + 2i)) (z - (-7 - 2i))

z² + bz + c = (z + (7 - 2i)) (z + (7 + 2i))

z² + bz + c = z² + ((7 - 2i) + (7 + 2i)) z + (7 - 2i) (7 + 2i)

z² + bz + c = z² + 14z + 53

so that b = 14 and c = 53, which makes b + c = 67.

3. If x + y = 4, then

x³ + y³ = 100

x³ + (4 - x)³ = 100

x³ + (64 - 48x + 12x² - x³) = 100

12x² - 48x + 64 = 100

12x² - 48x = 36

x² - 4x = 3

and by completing the square,

x² - 4x + 4 = 7

(x - 2)² = 7

x - 2 = ± 7

x = 2 + 7   or   x = 2 - 7

x = 9   or   x = -5

If x = 9, then y = -5, so one pair of solutions would be (x, y) = (9, -5).

nope.   happens to me as well.

-aparri