What is 1/2 x+1/3 y=2, 5/x-2y= 17/3 ?

The solution to 1/2 x+1/3 y=2, 5/x-2y= 17/3 is

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1/2 x+1/3 y=2, 5/x-2y= 17/3

\frac{1}{2} x + \frac{1}{3} y = 2, \frac{5}{x} - 2 y = \frac{17}{3}

(x = 4 + \frac{- 19 + 2269}{18}, x = 4 - \frac{19 + 2269}{18}, y = - \frac{- 19 + 2269}{12} y = \frac{19 + 2269}{12})

Solution steps

[\frac{1}{2} x + \frac{1}{3} y = 2 \frac{5}{x} - 2 y = \frac{17}{3}]

Isolate

x

for

\frac{1}{2} x + \frac{1}{3} y = 2 : x = 4 - \frac{2}{3} y

Plug the solutions

x = 4 - \frac{2}{3} y

into

\frac{5}{x} - 2 y = \frac{17}{3}

For

\frac{5}{x} - 2 y = \frac{17}{3}

, subsitute

x

with

4 - \frac{2}{3} y : y = - \frac{- 19 + 2269}{12}, y = \frac{19 + 2269}{12}

Plug the solutions

y = - \frac{- 19 + 2269}{12}, y = \frac{19 + 2269}{12}

into

\frac{1}{2} x + \frac{1}{3} y = 2

For

\frac{1}{2} x + \frac{1}{3} y = 2

, subsitute

y

with

- \frac{- 19 + 2269}{12} : x = 4 + \frac{- 19 + 2269}{18}

For

\frac{1}{2} x + \frac{1}{3} y = 2

, subsitute

y

with

\frac{19 + 2269}{12} : x = 4 - \frac{19 + 2269}{18}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

\frac{1}{2} x + \frac{1}{3} y = 2, \frac{5}{x} - 2 y = \frac{17}{3}

are

(x = 4 + \frac{- 19 + 2269}{18}, x = 4 - \frac{19 + 2269}{18}, y = - \frac{- 19 + 2269}{12} y = \frac{19 + 2269}{12})

x + 3 y = 1, 3 x^{2} + 3 y^{2} = 43

u + 6 = 3, v - 4 = - 2 u + v

y^{2} x + 5 y = 7, 3 x + 2 y = 5

x^{2} + 1^{2} = 25, z = 0

2 y + 100 = 2, x^{2} x = y + 250

What is 1/2 x+1/3 y=2, 5/x-2y= 17/3 ?
The solution to 1/2 x+1/3 y=2, 5/x-2y= 17/3 is