What value of $x$ $x$ satisfies the equation $\frac{7}{3} (x + \frac{9}{28}) = 20$ $7/3(x+9/28) = 20$ ?

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Answer 1 · 2016-11-07T17:25:53

$x = \frac{33}{4} = 8 \frac{1}{4}$ $x = 33/4=8 1/4$

The starting equation is:

$\frac{7}{3} (x + \frac{9}{28}) = 20$ $7/3(x+9/28)=20$

The first step is to multiply both sides by $3$ $3$ :

$7 (x + \frac{9}{28}) = 60$ $7(x+9/28)=60$

Now we divide both sides by $7$ $7$

$x + \frac{9}{28} = \frac{60}{7}$ $x+9/28=60/7$

$x = \frac{60}{7} - \frac{9}{28}$ $x=60/7-9/28$

$x = \frac{240}{28} - \frac{9}{28}$ $x=240/28-9/28$

$x = \frac{231}{28} = \frac{33}{4} = 8 \frac{1}{4}$ $x=231/28=33/4=8 1/4$