Solve the inequality both algebraically and graphically. Draw a number line graph of the solution and give interval notation.

Question

6xplus1less thanminusStartFraction 3 x Over 5 EndFraction
plus5
Question content area bottom
Part 1
The solution as an inequality is
  
enter your response here.
​(Type an inequality. Simplify your answer. Use integers or fractions for any numbers in the​ expression.)
Part 2
The solution in interval notation is
  
enter your response here.
​(Type your answer in interval notation. Use integers or fractions for any numbers in the​ expression.)
Part 3
Choose the correct graph of the solution below.
A.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 33 EndFraction. The region to the left of a right bracket at approximately 0.3 is shaded.
0
1
0
B.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 33 EndFraction. The region to the left of a right bracket at StartFraction 20 Over 33 EndFraction is shaded.
0
1
0
C.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 33 EndFraction. The region to the right of a left parenthesis at StartFraction 20 Over 33 EndFraction is shaded.
0
1
0
D.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 33 EndFraction. The region to the right of a left parenthesis at approximately 0.9 is shaded.
0
1
0
E.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 33 EndFraction. The region to the left of a right parenthesis at approximately 0.3 is shaded.
0
1
0
F.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 33 EndFraction. The region to the left of a right parenthesis at StartFraction 20 Over 33 EndFraction is shaded.
0
1
0
G.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 33 EndFraction. The region to the left of a right parenthesis at approximately 0.9 is shaded.
0
1
0

Answer 1

To solve the inequality \( 6x + 1 < -\frac{3x}{5} + 5 \), we will first simplify the inequality algebraically.

Part 1: Solve the Inequality

1. Start with the original inequality: \[ 6x + 1 < -\frac{3x}{5} + 5 \]

2. To eliminate the fraction, let's first multiply every term by 5 (the denominator): \[ 5(6x) + 5(1) < 5\left(-\frac{3x}{5}\right) + 5(5) \] Which simplifies to: \[ 30x + 5 < -3x + 25 \]

3. Now, we will move all terms involving \( x \) to one side and constant terms to the other side: \[ 30x + 3x < 25 - 5 \] Leading to: \[ 33x < 20 \]

4. Divide both sides by 33: \[ x < \frac{20}{33} \]

Thus, the solution to the inequality is: \[ x < \frac{20}{33} \]

Part 2: Interval Notation

In interval notation, this inequality can be written as: \[ (-\infty, \frac{20}{33}) \]

Part 3: Number Line Graph

We need to choose the appropriate graph for \( x < \frac{20}{33} \). The correct graph would show an open region to the left of \( \frac{20}{33} \) (since it is a "less than" inequality), indicating that all numbers less than \( \frac{20}{33} \) are included.

The only choices we have that represent this are:

• F: An infinite number line with the region shaded to the left of a right parenthesis at \( \frac{20}{33} \).

Summary of the Answers

1. The solution as an inequality is: \[ x < \frac{20}{33} \]
2. The solution in interval notation is: \[ (-\infty, \frac{20}{33}) \]
3. The correct graph of the solution is:
   • F: The region to the left of a right parenthesis at \( \frac{20}{33} \) is shaded.