How do you write the inequality and solve given "two thirds of a number plus 5 is greater than 12"?

Feb 13, 2017

The inequality is: $\frac{2}{3} n + 5 > 12$

The solution is: $n > \frac{21}{2}$

Explanation:

First, let's call the number we are going to solve for $n$.

Then two thirds of this number can be written as:

$\frac{2}{3} n$

Then if we "plus 5" or, in other words, add 5 we get:

$\frac{2}{3} n + 5$

Finally, if this is "greater than 12" we can finalize the inequality by writing:

$\frac{2}{3} n + 5 > 12$

Now, to solve this we first subtract $\textcolor{red}{5}$ from each side of the inequality to isolate the $n$ term while keeping the inequality balanced:

$\frac{2}{3} n + 5 - \textcolor{red}{5} > 12 - \textcolor{red}{5}$

$\frac{2}{3} n + 0 > 7$

$\frac{2}{3} n > 7$

Then, we multiply each side of the inequality by $\frac{\textcolor{red}{3}}{\textcolor{b l u e}{2}}$ to solve for $n$ while keeping the inequality balanced:

$\frac{\textcolor{red}{3}}{\textcolor{b l u e}{2}} \times \frac{2}{3} n > \frac{\textcolor{red}{3}}{\textcolor{b l u e}{2}} \times 7$

$\frac{\cancel{\textcolor{red}{3}}}{\cancel{\textcolor{b l u e}{2}}} \times \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{2}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} n > \frac{21}{2}$

$n > \frac{21}{2}$