Question

84. If 1/3 + 1/2 + 1/x = 4, then the value of x is​

Answer 1

GIVEN :

1/3 + 1/2 + 1/x = 4

TO FIND :

Value of x

SOLUTION :

1/3 + 1/2 + 1/x = 4

After LCM,

(2x + 3x + 6) / 6x = 4

(5x + 6)/6x = 4

5x + 6 = 24x

6 = 24 - 5x

6 = 19x

x = 6/19

Therefore, value of x is 6/19

Answer 2

Answer :-

6/19

Solution :-

$\sf \dfrac{1}{3} + \dfrac{1}{2} + \dfrac{1}{x} = 4$

Transpose 1/3 to RHS.

$\sf \dfrac{1}{2} + \dfrac{1}{x} = 4 - \dfrac{1}{3}$

Transpose 1/2 to RHS

$\sf \dfrac{1}{x} = 4 - \dfrac{1}{3} - \dfrac{1}{2}$

Taking LCM

$\sf \dfrac{1}{x} = \dfrac{4(6)}{1(6)} - \dfrac{1(2)}{3(2)} - \dfrac{1(3)}{2(3)}$

$\sf \dfrac{1}{x} = \dfrac{24}{6} - \dfrac{2}{6} - \dfrac{3}{6}$

$\sf \dfrac{1}{x} = \dfrac{24}{6} - \dfrac{5}{6}$

$\sf \dfrac{1}{x} = \dfrac{24 - 5}{6}$

$\sf \dfrac{1}{x} = \dfrac{19}{6}$

Reciprocal on both sides

$\sf \dfrac{x}{1} = \dfrac{6}{19}$

$\sf x = \dfrac{6}{19}$

Verification :-

$\sf \dfrac{1}{3} + \dfrac{1}{2} + \dfrac{1}{x} = 4$

$\sf \dfrac{1}{3} + \dfrac{1}{2} + \dfrac{1}{ \frac{6}{19} } = 4$

$\sf \dfrac{1}{3} + \dfrac{1}{2} + \dfrac{19}{6} = 4$

Taking LCM

$\sf \dfrac{1(2)}{3(2)} + \dfrac{1(3)}{2(3)} + \dfrac{19}{6} = 4$

$\sf \dfrac{2}{6} + \dfrac{3}{6} + \dfrac{19}{6} = 4$

$\sf \dfrac{2 + 3 + 19}{6} = 4$

$\sf \dfrac{24}{6} = 4$

$\sf 4 = 4$