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.
4/3y+2/5y<3/2Let's find the values for which any of the denominators is 0.
c|c
3y=0 & 5y=0
⇕ & ⇕
y=0 & y=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]
4/3y+2/5y < 3/2 & 4/3y+2/5y = 3/2
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*5*2y=30y.
We will divide the number line considering the excluded value y=0 and the solution to the related equation y= 5245.
Test Values
We will now test a value from each interval. Let's start with the interval y<0. We will test the point y=- 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
y< 0
y=- 1
-26< 22.5 ✓
0< y < 52/45
y=1/15
26<1.5 *
y>52/45
y=2
26<45 ✓
We found that the solutions to the given inequality are all the numbers less than 0 or larger than 5245.
Solution
y<0 or 52/45
