# How do you solve and write the following in interval notation: 1/2 x - 1/10 <=x + 1?

Apr 12, 2017

See the entire solution process below:

#### Explanation:

First, subtract $\textcolor{red}{\frac{1}{2} x}$ and $\textcolor{b l u e}{1}$ from each side of the equation to isolate the $x$ term while keeping the equation balanced:

$\frac{1}{2} x - \frac{1}{10} - \textcolor{red}{\frac{1}{2} x} - \textcolor{b l u e}{1} \le x + 1 - \textcolor{red}{\frac{1}{2} x} - \textcolor{b l u e}{1}$

$\frac{1}{2} x - \textcolor{red}{\frac{1}{2} x} - \frac{1}{10} - \textcolor{b l u e}{1} \le x - \textcolor{red}{\frac{1}{2} x} + 1 - \textcolor{b l u e}{1}$

$0 - \frac{1}{10} - \textcolor{b l u e}{\frac{10}{10}} \le \frac{2}{2} x - \textcolor{red}{\frac{1}{2} x} + 0$

$- \frac{11}{10} \le \frac{1}{2} x$

Now, multiply each side of the equation by $\textcolor{red}{2}$ to solve for $x$ while keeping the equation balanced:

$\textcolor{red}{2} \times - \frac{11}{10} \le \textcolor{red}{2} \times \frac{1}{2} x$

$- \frac{22}{10} \le \cancel{\textcolor{red}{2}} \times \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} x$

$- \frac{22}{10} \le x$

To state the solution in terms of $x$ we can reverse or "flip" the inequality:

$x \ge - \frac{22}{10}$

Or, in interval form:

$\left[- \frac{22}{10} , \infty\right)$