Question discussion

If 3x = 4y = 12-z, then $\left( {\frac{1}{x}\; + \;\frac{1}{y}\; + \;\frac{1}{z}} \right)$ is equal to:

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If 3^x = 4^y= 12^-z, then $\left( {\frac{1}{x}\; + \;\frac{1}{y}\; + \;\frac{1}{z}} \right)$ is equal to:

A0
B1
C3/2
D$- \frac{1}{2}$

Answer not set

Discussion 1

CCommunity11 May 2019

3x = 12-z

3 = 12-(z/x) -----equation 1

4y = 12-z

4 = 12-(z/y) -----equation 2

Multiplying $equation 1? and equation 2

121 = 12-{(z/x)+(z/y)}

1 = -{(z/x)+(z/y)}

1 = -(z/x) - (z/y)

1/z = -(1/x) - (1/y)

(1/x) + (1/y) + (1/z) = 0

If 3x = 4y = 12-z, then \(\left( {\frac{1}{x}\; + \;\frac{1}{y}\; + \;\frac{1}{z}} \right)\) is equal to:

Discussion 1