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Solve the inequality. Write your final answer in interval notation. $$ -2 x+3>x-5 $$

Short Answer

Expert verified
The solution is \((-\infty, \frac{8}{3})\).

Step by step solution

01

Simplify the inequality

Start by adding \(2x\) to both sides of the inequality to get all terms involving \(x\) on one side. This gives us:\[3 > 3x - 5\]
02

Isolate the variable term

Add 5 to both sides to move the constant on the right to the left side:\[3 + 5 > 3x\]which simplifies to:\[8 > 3x\]
03

Solve for x

To solve for \(x\), divide both sides by 3:\[\frac{8}{3} > x\]This can also be written as:\[ x < \frac{8}{3}\]
04

Write the solution in interval notation

Since \(x\) is less than \(\frac{8}{3}\), it means that \(x\) can be any number less than \(\frac{8}{3}\). Therefore, the solution in interval notation is:\[(-\infty, \frac{8}{3})\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Interval Notation
Interval notation is a way of writing subsets of the real number line using parentheses and brackets. It offers a compact method to express the solution set of an inequality. In the given example of solving the inequality \(-2x + 3 > x - 5\), we conclude that \(x < \frac{8}{3}\).The solution set includes all real numbers less than \(\frac{8}{3}\).
To express this in interval notation, we write it as \((-\infty, \frac{8}{3})\). Here's a quick guide:
  • Parentheses \(()\): Used when the endpoint is NOT included in the set. For example, \((a, b)\) means \(a < x < b\).
  • Brackets \([]\): Used when the endpoint IS included. For instance, \([a, b]\) means \(a \leq x \leq b\).
In our case, because \(x\) can be any number strictly less than \(\frac{8}{3}\), we use a parenthesis rather than a bracket next to \(\frac{8}{3}\). This ensures that \(\frac{8}{3}\) is not included in our solution set.
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and operators (such as \(+, -, \times, \div\)). They represent mathematical phrases to describe real-world situations or solve equations and inequalities.
In the exercise, the expression \(-2x + 3\) involves two components:
  • A variable term \(-2x\), which indicates that the variable \(x\) is multiplied by -2.
  • A constant, which in this case is \(+3\).
Understanding how to manipulate algebraic expressions is key to solving inequalities. In our exercise, adding or subtracting terms from both sides helped isolate the variable \(x\).
To solve inequalities successfully, it often involves:
  • Combining like terms
  • Using inverse operations to "move" terms from one side to another
  • Maintaining balance (whatever operation is done to one side is done to the other)
This foundational knowledge of algebraic expressions aids in transitioning to solving equations and inequalities.
Linear Inequality
Understanding linear inequalities is a fundamental aspect of algebra. A linear inequality is an inequality that involves a linear expression on one or both sides. Linear inequalities are akin to linear equations, but instead of equality, they use inequality symbols such as \(>, <, \geq,\text{ or } \leq\).
In the problem \(-2x + 3 > x - 5\), we have a linear inequality because both sides can be represented by linear expressions.
  • First, add \(2x\) to both sides to gather variable terms to one side, yielding \(3 > 3x - 5\).
  • Then, add \(5\) to both sides to simplify further, leading to \(8 > 3x\).
  • Finally, dividing both sides by \(3\) isolates \(x\), resulting in \(x < \frac{8}{3}\).
Working through each step follows the rule that any operation applied to one side of the inequality is also applied to the other side, much like solving equations. However, one must be cautious, as multiplying or dividing by a negative number requires flipping the inequality sign. Since we solved the inequality level by level, each step maintained a balance while isolating \(x\). Understanding and practicing these steps ensure you can solve any linear inequality you encounter.

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Most popular questions from this chapter

For the following exercises, solve the following polynomial equations by grouping and factoring. $$ 5 x^{3}+45 x=2 x^{2}+18 $$

For the following exercises, graph the function. Observe the points of intersection and shade the \(x\) -axis representing the solution set to the inequality. Show your graph and write your final answer in interval notation. $$ |x-2|<7 $$

For the following exercises, input the left-hand side of the inequality as a Y1 graph in your graphing utility. Enter \(\mathrm{Y} 2=\) the right-hand side. Entering the absolute value of an expression is found in the MATH menu, Num, liabs ( Find the points of intersection, recall \((2^{\text { nd }}\) CALC 5:intersection, lst curve, enter, } \(2^{\text { nd }}\) curve, enter, guess, enter). Copy a sketch of the graph and shade the \(x\) -axis for your solution set to the inequality. Write final answers in interval notation. $$ |x-4|<3 $$

For the following exercises, graph both straight lines (left-hand side being \(y_{1}\) and right-hand side being \(y_{2}\) ) on the same axes. Find the poin of intersection and solve the inequality by observing where it is true comparing the \(y\) -values of the lines. $$ \frac{1}{2} x+1>\frac{1}{2} x-5 $$

For the following exercises, input the left-hand side of the inequality as a Y1 graph in your graphing utility. Enter \(\mathrm{Y} 2=\) the right-hand side. Entering the absolute value of an expression is found in the MATH menu, Num, liabs ( Find the points of intersection, recall \((2^{\text { nd }}\) CALC 5:intersection, lst curve, enter, } \(2^{\text { nd }}\) curve, enter, guess, enter). Copy a sketch of the graph and shade the \(x\) -axis for your solution set to the inequality. Write final answers in interval notation. $$ |x+2|-5 < 2 $$

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