Question

XY/X+Y=20,YZ/Y+Z =40,ZX/Z+X=24​? Solve...Find X,Y,Z


Please Answer The Question

Answer 1

Answer:

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Answer 2

Values are $\bf x = 30, \: y = 60 , \: \text{and} \: z = 120$

Given:

• $\frac{xy}{x + y} = 20$
• $\frac{yz}{y + z} = 40$
• $\frac{zx}{z + x} = 24$

To find:

• Find the values of x,y and z.

Solution:

Concept to be used:

Reciprocal each equation and simplify.

Step 1:

Reverse each equation.

$\frac{x + y}{xy} = \frac{1}{20}$

or

$\frac{x}{xy} + \frac{y}{xy} = \frac{1}{20}$

or

$\frac{1}{y} + \frac{1}{x} = \frac{1}{20}$

or

$\bf \frac{1}{x} + \frac{1}{y} = \frac{1}{20} ...eq1$

by the same way

$\bf \frac{1}{y} + \frac{1}{z} = \frac{1}{40} ...eq2$

and

$\bf \frac{1}{x} + \frac{1}{z} = \frac{1}{24} ...eq3$

Step 2:

Add eq1, 2 and 3.

$\frac{1}{x} + \frac{1}{y} + \frac{1}{y} + \frac{1}{z} + \frac{1}{z} + \frac{1}{x} = \frac{1}{20} + \frac{1}{40} + \frac{1}{24}$

or

$2 \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) = \frac{1}{20} + \frac{1}{40} + \frac{1}{24}$

take LCM in RHS

LCM(20,40,24)=120

$2 \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) = \frac{6 + 3 + 5}{120}$

or

$\bf \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{7}{120}...eq4$

Step 3:

Subtract eq1 from eq4.

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} - \frac{1}{x} - \frac{1}{y} = \frac{7}{120} - \frac{1}{20}$

or

$\frac{1}{z}= \frac{7 - 6}{120}$

or

$\frac{1}{z}= \frac{1}{120}$

or

$\bf z = 120$

Step 4:

Subtract eq2 from eq4.

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} - \frac{1}{y} - \frac{1}{z} = \frac{7}{120} - \frac{1}{40}$

or

$\frac{1}{x} = \frac{7 - 3}{120}$

or

$\frac{1}{x} = \frac{4}{120}$

or

$\bf x = 30$

Step 5:

Subtract eq3 from eq4.

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} - \frac{1}{x} - \frac{1}{z} = \frac{7}{120} - \frac{1}{24}$

or

$\frac{1}{y} = \frac{7 - 5}{120}$

or

$\frac{1}{y} = \frac{2}{120}$

or

$\bf y = 60$

Thus,

Values are $\bf x = 30, \: y = 60 , \: \text{and} \: z = 120$

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