Question

3=(1/(1/8))+(1/z)

Answer 1

\[3=\frac{ 1 }{ \frac{ 1 }{ 8 }+\frac{ 1 }{ y } }\]

Answer 2

Multiply the numerator and denominator of the big fraction by 8y.

\( 3=\dfrac{ 1 }{ \frac{ 1 }{ 8 }+\frac{ 1 }{ y } } \times \dfrac{8y}{8y} \)

Answer 3

64z/65..?

Answer 4

@mathstudent55

Answer 5

@zepdrix

Answer 6

Got it nevermind, 24/5

Answer 7

\(3=\dfrac{ 1 }{ \frac{ 1 }{ 8 }+\frac{ 1 }{ y } } \times \dfrac{8y}{8y} \)

\(3=\dfrac{ 8y }{ \left(\frac{ 1 }{ 8 }+\frac{ 1 }{ y } \right) 8y } \)

\(3=\dfrac{ 8y }{ \frac{ 8y }{ 8 }+\frac{ 8y }{ y } } \)

\(3=\dfrac{ 8y }{ \frac{ \cancel8y }{ \cancel8 }+\frac{ 8\cancel y }{ \cancel y } } \)

\(3=\dfrac{ 8y }{ y +8} \)

\(3( y +8 )=8y \)

\(3y + 24 = 8y \)

\(24 = 5y\)

\(y = 4.8\)