# Negative three times a number plus four is no more than the number minus eight. What is the number?

Jun 28, 2018

number $n$ is such that $n \ge 3$

#### Explanation:

Let the number be $n$

Three times the number is $- 3 \times n$ = $3 n$

Plus four is $- 3 n + 4$

Number minus eight is $n - 8$

No more than is $\le$

So we get:

$- 3 n + 4 \le n - 8$

Simplify and solve this linear equation:

$- 3 n - n \le - 8 - 4$

$- 4 n \le - 12$

$n \ge - \frac{12}{-} 4$

$n \ge 3$

So the number $n$ is such that $n \ge 3$

Hope this helps!

Jun 28, 2018

See a solution process below:

#### Explanation:

"a number" $\to n$

"Negative three times a number" $\to - 3 n$

"Negative three times a number plus four" $\to - 3 n + 4$

"is no more than" $\to \le$ giving:

$- 3 n \le$

"the number minus eight" $\to n - 8$ giving:

$- 3 n + 4 \le n - 8$

Next, subtract $\textcolor{red}{n}$ and $\textcolor{b l u e}{4}$ from each side of the inequality to isolate the $n$ term while keeping the inequality balanced:

$- 3 n - \textcolor{red}{n} + 4 - \textcolor{b l u e}{4} \le n - \textcolor{red}{n} - 8 - \textcolor{b l u e}{4}$

$- 3 n - \textcolor{red}{1 n} + 0 \le 0 - 12$

$\left(- 3 - \textcolor{red}{1}\right) n \le - 12$

$- 4 n \le - 12$

Now, divide each side of the inequality by $\textcolor{b l u e}{- 4}$ to solve for $n$ while keeping the inequality balanced. However, because we are multiplying or dividing an inequality by a negative number we must reverse the inequality operator.

$\frac{- 4 n}{\textcolor{b l u e}{- 4}} \textcolor{red}{\ge} \frac{- 12}{\textcolor{b l u e}{- 4}}$

$\frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 4}}} n}{\cancel{\textcolor{b l u e}{- 4}}} \textcolor{red}{\ge} 3$

$n \ge 3$