To solve the given rational , we will follow four steps.
- State the excluded values. These are the values for which any of the denominators are 0.
- Solve the related .
- Use the values determined in the previous steps to divide a 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.
x/x+2+1/x-1<3/2
Let's find the values for which any of the denominators is 0.
c|c
x+2=0 & x-1=0
⇕ & ⇕
x=-2 & x=1The excluded values for this inequality are -2 and 1.
Solve the Related Equation
We obtain the related equation by replacing the inequality sign with an equals sign.
c|c
Inequality & Equation [0.8em]
x/x+2+1/x-1 < 3/2 & x/x+2+1/x-1 = 3/2
Let's solve the equation. First, we will get rid of all of the denominators. To do so, we will multiply both sides of the equation by 2(x+2)(x-1).
x/x+2+1/x-1=3/2
( x/x+2+1/x-1 ) 2(x+2)(x-1)=3/2(2)(x+2)(x-1)
x/x+2 * 2(x+2)(x-1) +1/x-1 * 2(x+2)(x-1)=3/2(2)(x+2)(x-1)
x/x+2 * 2 (x+2)(x-1) +1/x-1 * 2(x+2) (x-1)=3/2 (2)(x+2)(x-1)
2x(x-1)+2(x+2)=3(x-1)(x+2)
Now that we have an equation without any denominators, we can simplify it as much as possible.
2x(x-1)+2(x+2)=3(x-1)(x+2)
2x^2-2x+2x+4=(3x-3)(x+2)
2x^2-2x+2x+4=x(3x-3)+2(3x-3)
2x^2-2x+2x+4=3x^2-3x+6x-6
2x^2+4=3x^2+3x-6
- x^2+4=3x-6
- x^2-3x+4=-6
- x^2-3x+10=0
We obtained a . Let's identify the values of a, b, and c.
- x^2-3x+10=0 ⇕ -1x^2+( - 3)x+ 10=0
We have that a= - 1, b= - 3, and c= 10. We will solve this equation by substituting these values into the .
x=- b±sqrt(b^2-4ac)/2a
x=- ( - 3)±sqrt(( - 3)^2-4( -1)( 10))/2( -1)
x=3±sqrt((- 3)^2-4(- 1)(10))/2(-1)
x=3±sqrt(9+40)/-2
x=3±sqrt(49)/-2
x=3± 7/-2
We will find the solutions by dividing our answer in two using the positive and negative sign.
| x=3± 7/-2
|
| x=3- 7/-2
|
x=3+7/-2
|
| x=-4/-2
|
x=10/-2
|
| x=2
|
x=- 5
|
The solutions to the equation are x=2 and x=- 5.
Since the radicand is a negative number, the equation does not have solutions.
Divide the Number Line Into Intervals
We will divide the number line considering the excluded values x=- 2 and x=1 and the solutions to the related equation x=- 5 and x=2.
Test Values
We will now test a value from each interval. Let's start with the interval x<- 5. We will test the point x=- 6. 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.
x/x+2+1/x-1<3/2
- 6/-6+2+1/-6-1 ? < 3/2
- 6/-4+1/-7 ? < 3/2
6/4+1/- 7 ? < 3/2
6/4+( - 1/7 ) ? < 3/2
6/4 - 1/7 ? < 3/2
3/2-1/7 ? < 3/2
-1/7 ? < 0
- 1 < 0 ✓
Since we obtained a true statement, we can conclude that the numbers less than - 5 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
|
| x< - 5
|
x=- 6
|
- +1 < 0 ✓
|
| - 5< x < -2
|
x=-3
|
11<6 *
|
| -2
| x=0
|
- 2<3 ✓
|
| 1
| x=3/2
|
26<15 *
|
| x>2
|
x=3
|
11<15 ✓
|
We found that the solutions to the given inequality are all the numbers less than -5, between - 2 and 1, or greater than 2.
Solution
x<-5, -22
