1. ## Rational Inequalities

Solve each inequality by graphing, using technology. Express your answer to one decimal place.

a) 2x+1/x-3 <= x

Okay, I need help solving this. Am I on the right track?

1. Make equal to zero : x- 2x+1/x-3 <= 0

2. "combine terms"? : x(x-3)-2x+1 / x-3 <= 0
............................. x^2-5x+1 / x-3 ........up to here is this correct?

3. Now I look for the "roots" or "zeros" for numerator and denominator?
....N: x can't be ____?
....D: x can't be +3.

4. Put this (your roots/zeros) on a number line. Test numbers that are in between each root on the number line?
...f(-2) (a#less than 3) = .....
...f(4) (a# greater than 3)= .....

...Is this the right process? I've been having trouble with inequalities.

2. ## Re: Rational Inequalities

Originally Posted by tdotodot
3. Now I look for the "roots" or "zeros" for numerator and denominator?
....N: x can't be ____?
....D: x can't be +3.
Solve the numerator for 0 anyway. The values this gives are also "critical points" for your intervals. (They typically are points where the function goes from + to - or - to +.)

-Dan

3. ## Re: Rational Inequalities

Originally Posted by topsquark
Solve the numerator for 0 anyway. The values this gives are also "critical points" for your intervals. (They typically are points where the function goes from + to - or - to +.)

-Dan
Okay, here's what I did.

Numerator = 0 ->x=4.8 and x=0.2
Denominator = 0 -> x=3

Number line: |-----0.2-------3-------4.8-------|

Checked each number: 0.2 works , 3 does not work , 4.8 works.

Tested values in between each interval.
f(0.1) <=0
f(1) <=0
f(4) <=0
f(5) <=0

All integers tested between each interval were less than or equal to zero.

So then my final answer would be: (-infiniti,2) U (4, +infiniti)

Is that right?

4. ## Re: Rational Inequalities

Originally Posted by tdotodot
Solve each inequality by graphing, using technology. Express your answer to one decimal place.

a) 2x+1/x-3 <= x

Okay, I need help solving this. Am I on the right track?

1. Make equal to zero : x- 2x+1/x-3 <= 0
This is not quite right. You had $\dfrac{2x+1}{x-3} \le x$. Subtract $\dfrac{2x+1}{x-3}$ from each side to get $0 \le x - \dfrac{2x+1}{x-3}$. Your inequality is flipped the other way.

Originally Posted by tdotodot
2. "combine terms"? : x(x-3)-2x+1 / x-3 <= 0
............................. x^2-5x+1 / x-3 ........up to here is this correct?
Not quite. $x-\dfrac{2x+1}{x-3} = \dfrac{x(x-3) - (2x+1)}{x-3} = \dfrac{x^2-3x-2x-1}{x-3} = \dfrac{x^2-5x-1}{x-3}$. You had $x^2-5x+1$ in the numerator.

5. ## Re: Rational Inequalities

Originally Posted by SlipEternal
This is not quite right. You had $\dfrac{2x+1}{x-3} \le x$. Subtract $\dfrac{2x+1}{x-3}$ from each side to get $0 \le x - \dfrac{2x+1}{x-3}$. Your inequality is flipped the other way.

Not quite. $x-\dfrac{2x+1}{x-3} = \dfrac{x(x-3) - (2x+1)}{x-3} = \dfrac{x^2-3x-2x-1}{x-3} = \dfrac{x^2-5x-1}{x-3}$. You had $x^2-5x+1$ in the numerator.
Okay, so here is what I get now...

Numerator = 0 -> x=5.2 and x=-0.2
Denominator = 0 ->x=3

|--------(-0.2)-----------(3)----------(5.2)-----|

Test Values: (-0.2) = -0.0125 ... is Not greater than or equal to zero. *I am plugging this into "0 <= x^2-5x-1 / x-3"*
................... (3) = undefined
....................(5.2) = greater than or equal to zero.

Then pick values between each interval:
f(-1) -> 1.25 which is >= 0
f(1) -> 2.5 which is >= 0
f(4) -> -5 which is < 0
f(6) ->1.67 which is >= 0

So back to the original equation now?? ... 2x+1 / x-3 <= 0
.................................................. ......when is it less than or equal to zero?
.................................................. ......(4, 5.1)?
.................................................. ......I'm guessing the answer is wrong? Can someone please explain this to me?

6. ## Re: Rational Inequalities

Originally Posted by tdotodot
Solve each inequality by graphing, using technology. Express your answer to one decimal place.
a) 2x+1/x-3 <= x.
Look at this solution.