What is 1/x+1/y = 3/4 , 1/((6x))+1/((5y))= 2/15 ?

The solution to 1/x+1/y = 3/4 , 1/((6x))+1/((5y))= 2/15 is

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1/x+1/y = 3/4 , 1/((6x))+1/((5y))= 2/15

\frac{1}{x} + \frac{1}{y} = \frac{3}{4}, \frac{1}{( 6 x )} + \frac{1}{( 5 y )} = \frac{2}{15}

(x = 2, y = 4)

Solution steps

[\frac{1}{x} + \frac{1}{y} = \frac{3}{4} \frac{1}{( 6 x )} + \frac{1}{( 5 y )} = \frac{2}{15}]

Isolate

x

for

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

Plug the solutions

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

into

\frac{1}{( 6 x )} + \frac{1}{( 5 y )} = \frac{2}{15}

For

\frac{1}{( 6 x )} + \frac{1}{( 5 y )} = \frac{2}{15}

, subsitute

x

with

- \frac{4 y}{4 - 3 y} : y = 4

Plug the solutions

y = 4

into

\frac{1}{x} + \frac{1}{y} = \frac{3}{4}

For

\frac{1}{x} + \frac{1}{y} = \frac{3}{4}

, subsitute

y

with

4 : x = 2

Verify solutions by plugging them into the original equations

Therefore, the final solution for

\frac{1}{x} + \frac{1}{y} = \frac{3}{4}, \frac{1}{( 6 x )} + \frac{1}{( 5 y )} = \frac{2}{15}

(x = 2, y = 4)

y = 2 x + 5, y = x - 1

2019 x + 2020 y = 4025, 2020 x + 2019 x = 4023

2 x - y = 9, 2 x + y = 3

5 x - y = 9, 2 x + y = 5

2 x - 3 y = 1, 3 y = 3 x - 16

What is 1/x+1/y = 3/4 , 1/((6x))+1/((5y))= 2/15 ?
The solution to 1/x+1/y = 3/4 , 1/((6x))+1/((5y))= 2/15 is