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.
5/3a-3/4a>5/6Let's find the values for which any of the denominators is 0.
c|c
3a=0 & 4a=0
⇕ & ⇕
a=0 & a=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]
5/3a-3/4a > 5/6 & 5/3a-3/4a = 5/6
Let's solve the equation. First, we will get rid of all of the denominators by multiplying the equation by the least common denominator (LCD). The LCD of the given equation is 3* 4a=12a.
We will divide the number line considering the excluded value a=0 and the solution to the related equation a=1.1.
Test Values
We will now test a value from each interval. Let's start with the interval a<0. We will test the point a=- 1. If substituting this point into the original inequality produces a true statement, then the numbers in this interval are solutions. If not, these numbers are not solutions.
Since we obtained a true 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
a< 0
a=- 1
-11>10 *
0< a < 1.1
a=1
11>10 ✓
a>1.1
a=2
11>20 *
We found that the solutions to the given inequality are all the numbers between 0 and 1.1.
Solution
0
