How do you solve for D? d+7> $7 d + \frac{3}{3}$ $7d+3/3$

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Answer 1 · 2017-12-17T14:31:06

$d < 1$ $d<1$

We can treat this like an ordinary algebra problem!

$d + 7 > 7 d + \frac{3}{3}$ $d + 7 > 7d + 3/3$

The first thing to note is that $\frac{3}{3} = 1$ $3/3 = 1$

$\Rightarrow d + 7 > 7 d + 1$ $=> d+7>7d+1$

Subtracting $d$ $d$ on both sides:

$\Rightarrow 7 > 6 d + 1$ $=> 7 > 6d +1$

Subtracting $1$ $1$ on both sides:

$\Rightarrow 6 > 6 d$ $=> 6 > 6d$

Dividing by 6:

$\Rightarrow 1 > d$ $=> 1 > d$

$d < 1$ $d<1$

How do you solve for D? d+7>7d+337d+3/3