How do you solve \frac { 1} { 3} [ 2( x - 8) + 1] = 19?

Jun 9, 2018

$x = 36$

Explanation:

As per the question, we have

$\frac{1}{3} \left[2 \left(x - 8\right) + 1\right] = 19$

$\therefore \frac{1}{3} \left[2 x - 16 + 1\right] = 19$

$\therefore \frac{1}{3} \left[2 x - 15\right] = 19$

$\therefore \frac{1}{3} \left(2 x - 15\right) \times 3 = 19 \times 3$ ... [Multiplying $3$ on both sides]

$\therefore \frac{1}{\cancel{3}} \left(2 x - 15\right) \times \cancel{3} = 57$

$\therefore 2 x - 15 = 57$

$\therefore 2 x - 15 + 15 = 57 + 15$ ... [Adding $15$ on both sides]

$\therefore 2 x = 72$

$\therefore x = 36$

Jun 9, 2018

Here is an algebraic rewrite to make it simpler

Explanation:

First rewrite this in to simplest terms:
$\frac{1}{3} \left[2 \left(x - 8\right) + 1\right] = 19 \to 2 \left(x - 8\right) + 1 = \frac{19}{3}$,

$2 \left(x - 8\right) + 1 = \frac{19}{3} \to 2 \left(x - 8\right) = \frac{19}{3} - 1$,

$2 \left(x - 8\right) = \frac{19}{3} - 3 \to x - 8 = \frac{\frac{19}{3} - 1}{2}$,

$x - 8 = \frac{\frac{19}{3} - 1}{2} \to x = \frac{\frac{19}{3} - 1}{2} + 8$

Now you can evaluate the expression and solve for x.