Use the values determined in the previous steps to divide a number line into intervals.
Test a value in each interval to determine which intervals contain values that satisfy the inequality.
Let's deal with these steps one at a time.
Excluded Values
Consider the given inequality.
3/5x+1/6x>2/3
Let's find the values for which any of the denominators is x=0.
c|c
5x=0 & 6x=0
⇕ & ⇕
x=0 & x=0
The excluded value for this inequality is 0.
Solve the Related Equation
We obtain the related equation by replacing the inequality sign with an equals sign.
c|c
Inequality & Equation [0.8em]
3/5x+1/6x > 2/3 & 3/5x+1/6x = 2/3
Let's solve the equation. To do so, we will start by writing 6 as the product of 2 and 3 to help us finding the least common denominator (LCD).
3/5x+1/6x = 2/3
⇕
3/5 x+1/2( 3) x = 2/3
The least common denominator is the product of 5, x, 2, and 3.
Least Common Denominator
5( x)( 2)( 3) ⇔ 30x
To clear denominators, we will solve the equation by multiplying each side by the least common denominator.
We will divide the number line considering the excluded value x=0 and the solution to the related equation x= 2320.
Test Values
We will now test a value from each interval. Let's start with the interval x<0. We will test the value x=- 1. If this substitution produces a true statement, then the values in this interval are solutions. If not, these numbers are not solutions.
Since we obtained a false statement, we can conclude that the numbers less than 0 are not solutions to the inequality. Using the same process, we can determine whether the numbers that belong to the other intervals are solutions.
Interval
Test Value
Statement
x< 0
x=- 1
- 23≯ 20 *
0< x < 23/20
x=1
23> 20 ✓
x> 23/20
x=2
23≯40 *
We found that the solutions to the given inequality are all the numbers between 0 and 2320.
Solution
0
