Problem 46
Question
Solve each linear inequality in Exercises 27-48 and graph the solution set on a number line. Express the solution set using interval notation. $$\frac{4 x-3}{6}+2 \geq \frac{2 x-1}{12}$$
Step-by-Step Solution
Verified Answer
The solution of the inequality \(x \geq -\frac{19}{6}\). The solution set in interval notation is \([-19/6, \infty)\).
1Step 1: Simplify the equation
First, the given inequality \(\frac{4 x-3}{6}+2 \geq \frac{2 x-1}{12}\) should be simplified. In order to do this, all terms should be multiplied by 12, the least common multiple of the denominators, to remove the fractions. As a result, we get \(8x - 6 + 24 \geq 2x - 1\). Now combine the like terms, yielding \(8x + 18 \geq 2x - 1\) after simplifying.
2Step 2: Solve for x
Now, let's isolate x by subtracting 2x from both sides, resulting in \(6x + 18 \geq - 1\). Next, subtract 18 from both sides, to get \(6x \geq -19\). Finally, divide both sides by 6 to solve for x, ending up with \(x \geq -\frac{19}{6}\) as the solution for the inequality.
3Step 3: Write the solution in interval notation and on a number line
The solution in interval notation will be \([-19/6, \infty)\). For the number line, draw a closed dot (indicating inclusion) on -19/6 and draw a line extending to the right, towards positive infinity.
Key Concepts
Interval NotationInequality GraphingSolving Inequalities
Interval Notation
Interval notation is a way of representing solution sets for inequalities. When we solve an inequality like \( x \geq -\frac{19}{6} \), we often want to express the range of values that are included in the solution.
To do this succinctly, we use interval notation. The solution \( x \geq -\frac{19}{6} \) includes all numbers greater than or equal to \(-\frac{19}{6}\).
To do this succinctly, we use interval notation. The solution \( x \geq -\frac{19}{6} \) includes all numbers greater than or equal to \(-\frac{19}{6}\).
- The square bracket \([ \) means "including" the endpoint. So \([-\frac{19}{6}, \infty) \) indicates that the range starts at \(-\frac{19}{6}\) and goes up to infinity, including \(-\frac{19}{6}\) itself.
- In contrast, a parenthesis \(( \) would mean the endpoint is not included. For example, if the symbol were a parenthesis, \((-\frac{19}{6}, \infty) \) would mean \(-\frac{19}{6}\) is not part of the solution set.
Inequality Graphing
Graphing linear inequalities provides a visual representation of solutions on a number line.
Let's break down how you would do this:
Let's break down how you would do this:
- First, draw a number line. This line acts as your visual framework.
- Locate the key value from your inequality; in this case, \(-\frac{19}{6}\).
- Since our inequality is \( x \geq -\frac{19}{6} \), this means \(-\frac{19}{6}\) is included. Represent this with a filled (or closed) dot on the number line.
- With the dot marking \(-\frac{19}{6}\), draw a line extending to the right to show all the numbers greater than \(-\frac{19}{6}\). This indicates that every number in that direction satisfies the inequality.
Solving Inequalities
Solving linear inequalities is similar to solving linear equations, but with additional rules when multiplying or dividing by negative numbers. In the problem \( \frac{4x-3}{6} + 2 \geq \frac{2x-1}{12} \), the aim is to isolate the variable \(x\).
Here’s a simplified guide to solving such inequalities:
Here’s a simplified guide to solving such inequalities:
- Remove fractions: Multiply all terms by the least common multiple of the denominators. This eliminates fractions and simplifies subsequent steps. In this case, multiplying by 12 removes the denominators.
- Simplify: Combine like terms and simplify each side as you would in an equation.
- Isolate \(x\): Use basic algebraic operations to get \(x\) alone. This includes moving all terms involving \(x\) to one side and constants to the opposite side, just as you saw where \(6x + 18 \geq -1\) was simplified to \(6x \geq -19\).
- Consider the direction: If you multiply or divide by a negative number, reverse the inequality sign. This step is crucial and different from solving equations.
Other exercises in this chapter
Problem 46
Solve each rational inequality in Exercises \(29-48,\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \fra
View solution Problem 46
Solve each equation in Exercises \(39-54\) by completing the square. $$ x^{2}+6 x-5=0 $$
View solution Problem 46
Exercises \(31-50\) contain equations with variables in denominators. For each equation, a. Write the value or values of the variable that make a denominator ze
View solution Problem 47
Explain how to multiply complex numbers and give an example.
View solution