# How do you solve the inequality: 5x + 10 >=10 and 7x - 7 <=14?

Apr 15, 2018

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Required Solution: color(blue)(0<=x<=3

Interval Notation: color(blue)([ 0,3 ]

#### Explanation:

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The Compound Inequality Expression given:

color(red)(5x+10>=10 and 7x-7<=14

color(green)("Step 1"

Consider $5 x + 10 \ge 10$ first and simplify.

Subtract color(red)(10 to both sides of the inequality to obtain

$5 x + 10 - \textcolor{red}{10} \ge 10 - \textcolor{red}{10}$

$5 x + \cancel{10} - \textcolor{red}{\cancel{10}} \ge \cancel{10} - \textcolor{red}{\cancel{10}}$

$5 x \ge 0$

Divide both sides of the inequality by $\textcolor{red}{5}$

$\frac{5 x}{\textcolor{red}{5}} \ge \frac{0}{\textcolor{red}{5}}$

$\frac{\cancel{5} x}{\textcolor{red}{\cancel{5}}} \ge \frac{0}{\textcolor{red}{5}}$

color(blue)(x>=0 Intermediate Solution 1

color(green)("Step 2"

Consider $7 x - 7 \le 14$ next.

Add $\textcolor{red}{7}$ to both sides of the inequality to get

$7 x - 7 + \textcolor{red}{7} \le 14 + \textcolor{red}{7}$

$7 x - \cancel{7} + \textcolor{red}{\cancel{7}} \le 14 + \textcolor{red}{7}$

$7 x \le 21$

Divide both sides of the inequality by color(red)(7

(7x)/color(red)(7)<=21/color(red)(7

(cancel 7x)/color(red)(cancel 7)<=cancel 21^color(red)3/color(red)(cancel 7

color(blue)(x<=3 Intermediate Solution 2

color(green)("Step 3"

Combine both the Intermediate Solution to obtain:

color(blue)(x>=0 and color(blue)(x<=3

color(red)(0<=x<=3 The required solution

Using the interval notation: color(red)([0,3]

color(green)("Step 4"

You can verify the results using a graph

The first graph below is created using the two inequality expressions given:

color(red)(5x+10>=10 and 7x-7<=14

The solution is the shaded common region

You can also graph just the solution to obtain a solution graph:

color(red)(0<=x<=3

Hope you find this solution useful.