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

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Answer 1 · 2018-06-09T07:03:05

$x = 36$ $x=36$

As per the question, we have

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

$:.1/3[2x-16+1]=19$

$:.1/3[2x-15]=19$

$:.1/3(2x-15)xx3=19xx3$ ... [Multiplying $3$ on both sides]

$:.1/cancel3(2x-15)xxcancel3=57$

$:.2x-15=57$

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

$:.2x=72$

$:.x=36$

Hence, the answer.

Answer 2 · 2018-06-09T07:07:33

Here is an algebraic rewrite to make it simpler

First rewrite this in to simplest terms:
$1/3[2(x-8)+1]=19 -> 2(x-8)+1 = 19/3$ ,

$2(x-8)+1=19/3 ->2(x-8)= 19/3-1$ ,

$2(x-8)=19/3 -3 -> x-8=(19/3-1)/2$ ,

$x-8=(19/3-1)/2 -> x=(19/3-1)/2 +8$

Now you can evaluate the expression and solve for x.

How do you solve \frac { 1} { 3} [ 2( x - 8) + 1] = 1913[2(x−8)+1]=19\frac { 1} { 3} [ 2( x - 8) + 1] = 19?