What is (1/x-y)+(1/x+y)=6,(1/x-y)-(1/x+y)=4 ?

The solution to (1/x-y)+(1/x+y)=6,(1/x-y)-(1/x+y)=4 is

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(1/x-y)+(1/x+y)=6,(1/x-y)-(1/x+y)=4

(\frac{1}{x} - y) + (\frac{1}{x} + y) = 6, (\frac{1}{x} - y) - (\frac{1}{x} + y) = 4

(x = \frac{1}{3}, y = - 2)

Solution steps

[(\frac{1}{x} - y) + (\frac{1}{x} + y) = 6 (\frac{1}{x} - y) - (\frac{1}{x} + y) = 4]

Solve

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

For

(\frac{1}{x} - y) + (\frac{1}{x} + y) = 6

, subsitute

y

with

- 2

\frac{1}{x} - (- 2) + \frac{1}{x} - 2 = 6

\frac{1}{x} - (- 2) + \frac{1}{x} - 2 = 6 : x = \frac{1}{3}

Verify solutions by plugging them into the original equations

Therefore, the final solution for

(\frac{1}{x} - y) + (\frac{1}{x} + y) = 6, (\frac{1}{x} - y) - (\frac{1}{x} + y) = 4

(x = \frac{1}{3}, y = - 2)

2 x - y = 4, - 6 x + 3 y = 18

(1) 2 x + 3 y = 1, 3 x - y = 7

\frac{3}{x} + \frac{2}{y} = \frac{9}{x} \cdot y, \frac{9}{x} + \frac{4}{y} = \frac{21}{x} \cdot y

3 x + y = 2, 3 x + 2 y = 7

(x - 7)^{2} = 3 y, x = 0

What is (1/x-y)+(1/x+y)=6,(1/x-y)-(1/x+y)=4 ?
The solution to (1/x-y)+(1/x+y)=6,(1/x-y)-(1/x+y)=4 is