# How do you solve 5 - x > 2(x+1)?

Aug 31, 2015

$x \in \left(- \infty , 1\right)$

#### Explanation:

Your goal here is to isolate $x$ on one side of the inequality. Start by suing the distributive property of multiplication to expand the paranthesis on the right-hand side

$5 - x > 2 \cdot x + 2 \cdot 1$

Move $2 x$ on the left-hand side of the inequality, and $5$ on the right-hand side of the inequality - do not forget to change their signs!

$- x - 2 x > 2 - 5$

$- 3 x > - 3$

Finally, divide both sides by $\left(- 3\right)$, but keep in mind that you need to change the sign of the inequality as well

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 3}}} \cdot x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 3}}}} < \frac{\left(- 3\right)}{\left(- 3\right)}$

$x < 1$

So, for any value of $x < 1$, the inequality will be true. The solution set will thus be $x \in \left(- \infty , 1\right)$.
The text contains a mathematical question in the form of an inequality and provides a detailed solution to it.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# How do you solve 5 - x > 2(x+1)?

Aug 31, 2015

$x \in \left(- \infty , 1\right)$

#### Explanation:

Your goal here is to isolate $x$ on one side of the inequality. Start by suing the distributive property of multiplication to expand the paranthesis on the right-hand side

$5 - x > 2 \cdot x + 2 \cdot 1$

Move $2 x$ on the left-hand side of the inequality, and $5$ on the right-hand side of the inequality - do not forget to change their signs!

$- x - 2 x > 2 - 5$

$- 3 x > - 3$

Finally, divide both sides by $\left(- 3\right)$, but keep in mind that you need to change the sign of the inequality as well

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 3}}} \cdot x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 3}}}} < \frac{\left(- 3\right)}{\left(- 3\right)}$

$x < 1$

So, for any value of $x < 1$, the inequality will be true. The solution set will thus be $x \in \left(- \infty , 1\right)$.
user
The text contains a mathematical question in the form of an inequality and provides a detailed solution to it.

TYPE: SINGLE_QUESTION_WITH_ANSWER