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.
1/3b+1/4b<1/5Let's find the values for which any of the denominators is 0.
c|c
3b=0 & 4b=0
⇕ & ⇕
b=0 & b=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]
1/3b+1/4b < 1/5 & 1/3b+1/4b = 1/5
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*4*5b=60b.
We will divide the number line considering the excluded value b=0 and the solution to the related equation b= 3512.
Test Values
We will now test a value from each interval. Let's start with the interval b<0. We will test the point b=- 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 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
b< 0
b=- 1
-7< 2.4 ✓
0< b < 35/12
b=1/12
7<0.2 *
b>35/12
b=3
7<7.2 ✓
We found that the solutions to the given inequality are all the numbers less than 0 or larger than 3512.
Solution
b<0 or 3512
