# Solve for x in |(3+x,5,2),(11-3x,17,16),(7-x,14,13)| = 0?

Aug 6, 2018

$x = \frac{47}{31}$

#### Explanation:

Here ,

$| \left(3 + x , 5 , 2\right) , \left(11 - 3 x , 17 , 16\right) , \left(7 - x , 14 , 13\right) | = 0$

Taking ${R}_{2} + 3 {R}_{1} \mathmr{and} {R}_{3} + {R}_{1}$

$| \left(3 + x , 5 , 2\right) , \left(11 - 3 x + 9 + 3 x , 17 + 15 , 16 + 6\right) , \left(7 - x + 3 + x , 14 + 5 , 13 + 2\right) | = 0$

$\therefore | \left(3 + x , 5 , 2\right) , \left(20 , 32 , 22\right) , \left(10 , 19 , 15\right) | = 0$

Taking ${C}_{2} - {C}_{3}$

$\therefore | \left(3 + x , 3 , 2\right) , \left(20 , 10 , 22\right) , \left(10 , 4 , 15\right) | = 0$

Taking ${C}_{1} - {C}_{2}$

$\therefore | \left(x , 3 , 2\right) , \left(10 , 10 , 22\right) , \left(6 , 4 , 15\right) | = 0$

Expanding we get

$x \left(10 \cdot 15 - 4 \cdot 22\right) - 3 \left(10 \cdot 15 - 6 \cdot 22\right) + 2 \left(10 \cdot 4 - 6 \cdot 10\right)$=$0$

$\therefore x \left(150 - 88\right) - 3 \left(150 - 132\right) + 2 \left(40 - 60\right) = 0$

$\therefore 62 x - 54 - 40 = 0$

$\therefore 62 x = 94$

$\therefore x = \frac{47}{31}$