What is (125^x)/((5^y))=25,2^{3x}*(1/8)^{1-y}=32 ?

The solution to (125^x)/((5^y))=25,2^{3x}*(1/8)^{1-y}=32 is

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(125^x)/((5^y))=25,2^{3x}*(1/8)^{1-y}=32

\frac{12 5 ^{x}}{( 5 ^{y} )} = 25, 2^{3 x} \cdot (\frac{1}{8})^{1 - y} = 32

(x = \frac{7}{6}, y = \frac{3}{2})

Solution steps

[\frac{12 5 ^{x}}{( 5 ^{y} )} = 25 2^{3 x} (\frac{1}{8})^{1 - y} = 32]

Isolate

x

for

\frac{12 5 ^{x}}{( 5 ^{y} )} = 25 : x = \frac{2 + y}{3}

Plug the solutions

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

into

2^{3 x} \cdot (\frac{1}{8})^{1 - y} = 32

For

2^{3 x} \cdot (\frac{1}{8})^{1 - y} = 32

, subsitute

x

with

\frac{2 + y}{3} : y = \frac{3}{2}

Plug the solutions

y = \frac{3}{2}

into

\frac{12 5 ^{x}}{( 5 ^{y} )} = 25

For

\frac{12 5 ^{x}}{( 5 ^{y} )} = 25

, subsitute

y

with

\frac{3}{2} : x = \frac{7}{6}

Verify solutions by plugging them into the original equations

Therefore, the final solution for

\frac{12 5 ^{x}}{( 5 ^{y} )} = 25, 2^{3 x} \cdot (\frac{1}{8})^{1 - y} = 32

(x = \frac{7}{6}, y = \frac{3}{2})

\frac{41}{x} - \frac{1}{( y - 2 )} = - 1, \frac{4}{x} + \frac{3}{( y - 2 )} = 5

x + y = 7, 2 x + 3 y = 8

3 x - 2 y = 1, 3 x + 5 y = 13

- 8 + x + 8 x = 2, y = 2

3 x + y + z = 1, x + 2 y - z = 1, x + y + 2 z = 17

What is (125^x)/((5^y))=25,2^{3x}*(1/8)^{1-y}=32 ?
The solution to (125^x)/((5^y))=25,2^{3x}*(1/8)^{1-y}=32 is