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Q21.

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Found in: Page 302

### Algebra 1

Book edition Student Edition
Author(s) Carter, Cuevas, Day, Holiday, Luchin
Pages 801 pages
ISBN 9780078884801

# Solve each inequality. Check your solution. $2x+11\le 5x-10$

The required value of x is $\left[7,\infty \right)$.

See the step by step solution

## Step 1. State the concept for inequality.

To graph the endpoint with strict inequality < or >, use the open parenthesis or a hollow circle at that endpoint.

To graph the endpoint with inequality symbol $\le \text{or}\ge$, use the bracket or a solid circle at that endpoint.

Properties of inequality:

$1.\text{}b\le c⇒b±a\le c±a\phantom{\rule{0ex}{0ex}}\begin{array}{l}2.\text{}b\le c⇒ab\le ac,\text{if}a>0\\ 3.\text{}b\le c⇒ab\ge ac,\text{if}a<0\end{array}$

## Step 2. Solve the inequality for x.

In order to solve the inequality:

$2x+11\le 5x-10$

Subtract both sides of the inequality by 2x, as

$\begin{array}{c}2x+11\le 5x-10\\ 2x+11-2x\le 5x-10-2x\\ 11\le 3x-10\\ 11+10\le 3x-10+10\\ 21\le 3x\\ \frac{21}{3}\le \frac{3x}{3}\\ 7\le x\\ x\ge 7\end{array}$

Thus, the inequality is true for x greater than or equal to 7, thus the solution of given inequality in interval form is, as $\left[7,\infty \right)$.

## Step 3. Check the solution.

To check the answer solution for the given inequality.

First check the solution at end point by converting it into an equation.

If the solution is less than m, then pick any value of x less than m and check whether it satisfies the inequality.

If the solution is greater than m, then pick any value of x greater than m and check whether it satisfies the inequality.

Now, to check the answer solution for the given inequality first check the solution at end point by converting it into an equation. Putting x equals to 7 in the inequality as

$\begin{array}{c}2\left(7\right)+11\le 5\left(7\right)-10\\ 25\le 25\\ \text{Left Hand side}=\text{Right Hand side}\end{array}$

Now put any value any value of x greater than 7, say 9 and check whether it satisfies the inequality, as

$\begin{array}{c}2\left(9\right)+11\le 5\left(9\right)-10\\ 18+11\le 45-10\end{array}\phantom{\rule{0ex}{0ex}}29\le 35\text{}\left[\text{True}\right]$

Since, both conditions are satisfied thus solution of inequality is correct.