How do you solve $\frac{j}{4} - 8 < 4$ $j/4 - 8 < 4$ ?

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Answer 1 · 2018-03-22T03:20:39

$j < 48$ $j<48$

You must isolate $j$ $j$ algebraically.

$\frac{j}{4} - 8 < 4$ $j/4-8<4$

Your first step in solving this inequality is to add $8$ $8$ to both sides.

$\frac{j}{4} < 12$ $j/4<12$

Now multiply both sides by $4$ $4$ to get rid of the denominator.

$j < 48$ $color(blue)(j<48)$

So for any number, $j$ $j$ , that is less than $48$ $48$ , the inequality is true.
You can check this by plugging in values.

$j = 52$ $j=52$ , then $\frac{52}{4} - 8 < 4$ $52/4-8<4$

$\Rightarrow 13 - 8 < 4$ $=> 13-8<4$

$\Rightarrow 5 < 4$ $=> 5<4$

This is not true, strengthening the statement. Feel free to plug in other values.

How do you solve j4−8<4j/4 - 8 < 4?