Question

In the system of equation $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=6,\frac{2}{x}-\frac{3}{y}+\frac{4}{z}=8$ and $\frac{3}{x}-\frac{4}{y}+\frac{5}{z}=10$, values of $x$, $y$ and $z$ which satisfy the above three equations, will be

• A. $\frac{1}{2}$, 1 and $\frac{1}{3}$
• B. $\frac{1}{3}$, 1 and $\frac{1}{2}$
• C. $1,\frac{1}{2}$ and $\frac{1}{3}$
• D. $1,\frac{1}{3}$ and $\frac{1}{2}$

Answer 1

The correct option is C $1,\frac{1}{2}$ and $\frac{1}{3}$

If $\frac{1}{x}=A,\frac{1}{y}=B$ and $\frac{1}{z}=C$, then the given equation will be written as:

$A+B+C=6$.........(i)

$2A-3B+4C=8$ ...............(ii)

$3A-4B+5C=10$ ..... (iii)

On multiplying equation (i) by 3 and adding it to equation (ii),

$(3A+3B+3C)+(2A-3B+4C)=6\times 3\times 8$

$5A+7C=26$ ...... (iv)

On multiplying equation (i) by 4 and adding it to equation (iii),

$(4A+4B+4C)+(3A-4B+5C)=6\times 4+10$

$7A+9C=34$ ..... (v)

On multiplying equation (iv) by 7 and equation(v) by 5 and subtracting equation (v) from equation (iv),

$(35A+49C)-(35A+45C)=26\times 7-34\times 5$

$4C=182-170=12\Rightarrow C=\frac{12}{4}=3$

On substituting this value of C in equation (iv),

$5a+7\times 3=26\Rightarrow 5A=26-21\therefore 5A=5$

$\Rightarrow A=\frac{5}{5}=1$

From equation (i), $1+B+3=6\Rightarrow B=6-4=2$

$\because \frac{1}{x}=A=1\Rightarrow x=1,\frac{1}{y}=B=2\Rightarrow y=\frac{1}{2}$

and $\frac{1}{z}=C=3\Rightarrow z=\frac{1}{3}$

Hence values of x, y and z will be $1,\frac{1}{2}$ and $\frac{1}{3}$ respectively.