Question

Solve:
$\frac{1}{x}-\frac{2}{y}+4=0$

$\frac{1}{y}-\frac{1}{z}+1=0$

$\frac{2}{z}+\frac{3}{x}=14$

• A. $x=2y=3z=4$
• B. $x=\frac{1}{2}y=\frac{1}{3}z=\frac{1}{4}$
• C. $x=\frac{1}{4}y=\frac{1}{2}z=\frac{1}{3}$
• D. $x=\frac{1}{2}y=\frac{1}{3}z=\frac{1}{2}$

Answer 1

The correct option is B $x=\frac{1}{2}y=\frac{1}{3}z=\frac{1}{4}$

$\frac{1}{x}-\frac{2}{y}+4=0$----------(1)
$\frac{1}{y}-\frac{1}{z}+1=0$-----------(2)
$\frac{2}{z}+\frac{3}{x}=14$------------(3)
Let
$\frac{1}{x}=A,\frac{1}{y}=B,\frac{1}{z}=C$
The equations become,
$A-2B=-4-----\left(4\right)B-C=-1------\left(5\right)2C+3A=14---\left(6\right)$
$eqn\left(4\right)+eqn\left(5\right)\times 2$ gives
A - 2B = -4
2B -2C = -2
______________
A - 2C = -6 ----------(7)

$eqn\left(6\right)+eqn\left(7\right)$ gives
3A + 2C = 14
A -2C = -6
______________
4A = 8
A=2----------(8)
Substituteing $eqn\left(8\right)$ in $eqn\left(4\right)$ we get,

2 - 2B = -4
-2B = -6
B = 3 ----------------(9)
Substituting eqn(9)\) in eqn(5)\) we get,
3 - C = -1
-C = -4
C = 4 ---------------(10)

$A=\frac{1}{x}=2$
$x=\frac{1}{2}$

$B=\frac{1}{y}=3$
$y=\frac{1}{3}$

$C=\frac{1}{z}=4$
$z=\frac{1}{4}$
Verification :
Substituting the value of x and y in eqn(1) we get

$\frac{1}{\frac{1}{2}}-\frac{2}{\frac{1}{3}}=2-6+4=-4+4=0=RHS$
Substituting the values of y and z in eqn(2) we get,
$\frac{1}{\frac{1}{3}}-\frac{1}{\frac{1}{4}}+1$
$=3-4+1$
$4-4$
$0=RHS$