What is xy=1,2(x-1)(y+2)=15 ?

The solution to xy=1,2(x-1)(y+2)=15 is

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xy=1,2(x-1)(y+2)=15

x y = 1, 2 (x - 1) (y + 2) = 15

(x = - \frac{- 17 + 321}{8}, x = \frac{321 + 17}{8}, y = - \frac{17 + 321}{4} y = \frac{321 - 17}{4})

Solution steps

[x y = 1 2 (x - 1) (y + 2) = 15]

Isolate

x

for

x y = 1 : x = \frac{1}{y}; y \neq = 0

Plug the solutions

x = \frac{1}{y}

into

2 (x - 1) (y + 2) = 15

For

2 (x - 1) (y + 2) = 15

, subsitute

x

with

\frac{1}{y} : y = - \frac{17 + 321}{4}, y = \frac{321 - 17}{4}

Plug the solutions

y = - \frac{17 + 321}{4}, y = \frac{321 - 17}{4}

into

x y = 1

For

x y = 1

, subsitute

y

with

- \frac{17 + 321}{4} : x = - \frac{- 17 + 321}{8}

For

x y = 1

, subsitute

y

with

\frac{321 - 17}{4} : x = \frac{321 + 17}{8}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

x y = 1, 2 (x - 1) (y + 2) = 15

are

(x = - \frac{- 17 + 321}{8}, x = \frac{321 + 17}{8}, y = - \frac{17 + 321}{4} y = \frac{321 - 17}{4})

y = x^{2} - 2 x + 3, y = - 5 x + 1

2 x + 5 \frac{y}{x} y = 6, 4 x - 5 \frac{y}{x} y = - 3

3 x + 5 y = 3, 2 x^{1} y = - 5

x - 1 = 1, x^{3} - y^{3} = 7

y = 2 x^{2} + z^{2}, y = 6 - 2 x^{2} - z^{2}