non linear inequalities with fractions

[abel muroi]

I have a question about inequalities with fractions.

I was given the problem -2 < (x+1)/(x - 3) and was told to put the solution in interval notation.

My question is, if i were to subtract (x +1)/(x - 3) from both sides.. does that mean that all the signs in the fraction become negative?

for example (-x - 1)/(-x + 3) - 2 < 0

is that how you subtract fractions from both sides?

 

[pka]

I have a question about inequalities with fractions.
I was given the problem -2 < (x+1)/(x - 3) and was told to put the solution in interval notation.

If I were you, I would solve $\displaystyle \dfrac{x+1}{x-3}+2>0$.

 

[abel muroi]

If I were you, I would solve $\displaystyle \dfrac{x+1}{x-3}+2>0$.

Oh I see what you did there. So it's pretty much the same if i were to just flip the inequality in the reverse order as long as i switch the greater than sign.

Thanks for the tip!

 

[HallsofIvy]

If I were you, I would solve $\displaystyle \dfrac{x+1}{x-3}+2>0$.

I wouldn't. If $\displaystyle \dfrac{x+1}{x-3}> -2$ then there are two cases:
1) x- 3> 0. Then x+ 1> -2(x- 3)= -2x+ 6. Add 2x to both sides and subtract 1 from both sides. Make sure the answer here also satisfies x> 3.

2) x- 3< 0. Then x+ 1< -2(x- 3)= -2x+ 6. Again add 2x to both sides and subtract 6 from both sides. Make sure the answer here also satisfies x< 3.

 

[Steven G]

I wouldn't. If $\displaystyle \dfrac{x+1}{x-3}> -2$ then there are two cases:
1) x- 3> 0. Then x+ 1> -2(x- 3)= -2x+ 6. Add 2x to both sides and subtract 1 from both sides. Make sure the answer here also satisfies x> 3.

2) x- 3< 0. Then x+ 1< -2(x- 3)= -2x+ 6. Again add 2x to both sides and subtract 6 from both sides. Make sure the answer here also satisfies x< 3.

This of course is the way to do these problems. The thing is that Abel does not want to learn how to solve inequalities with And in it. Abel like to partition the number line and use test points.

 

[Steven G]

I have a question about inequalities with fractions.

I was given the problem -2 < (x+1)/(x - 3) and was told to put the solution in interval notation.

My question is, if i were to subtract (x +1)/(x - 3) from both sides.. does that mean that all the signs in the fraction become negative?

for example (-x - 1)/(-x + 3) - 2 < 0

is that how you subtract fractions from both sides?

You should note that (-x - 1)/(-x + 3) =[-1(x+1)]/[-1(x-3)] = (x+1)/(x-3). So what you did was move (x+1)/(x-3) to the other side without changing the sign. Fine what you wrote looked different on the other side but it really was the same. You should subtract (x+1)/(x-3) from both sides. The the left side becomes -(x+1)/(x-3) - 2< 0 or (x+1)/(x-3) + 2< 0

 

[Steven G]

Oh I see what you did there. So it's pretty much the same if i were to just flip the inequality in the reverse order as long as i switch the greater than sign.

Thanks for the tip!

Not really. -2 < (x+1)/(x - 3) means that (x+1)/(x - 3) > -2 or (x+1)/(x - 3) + 2 > 0 ....

The logic is like this. A < B because B > A

 

[pka]

If I were you, I would solve $\displaystyle \dfrac{x+1}{x-3}+2>0$.

I wouldn't. If $\displaystyle \dfrac{x+1}{x-3}> -2$ then there are two cases:

Well I do like the easy way: See here. Look at [h=2]Alternate forms:[/h]

 

[abel muroi]

This of course is the way to do these problems. The thing is that Abel does not want to learn how to solve inequalities with And in it. Abel like to partition the number line and use test points.

Yep, you know me pretty well already

 

[Dale10101]

Also

I have a question about inequalities with fractions.

I was given the problem -2 < (x+1)/(x - 3) and was told to put the solution in interval notation.

My question is, if i were to subtract (x +1)/(x - 3) from both sides.. does that mean that all the signs in the fraction become negative?

for example (-x - 1)/(-x + 3) - 2 < 0

is that how you subtract fractions from both sides?

Reversal Property

(You can swap a and b over, if you make sure the inequality symbol still "points at" the smaller value.)

If a > b then b < a
If a < b then b > a

Addition and Subtraction property

(in all cases you can add or subtract the same value to or from each side without affecting the inequality sign)

If a < b, then a + c < b + c
If a < b, then a − c < b − c
If a > b, then a + c > b + c, and
If a > b, then a − c > b – c

Multiplication and Division property

(when you multiply both sides by a negative number then reverse the inequality symbol.)

If a < b, and c is positive, then ac < bc
If a < b, and c is negative, then ac > bc (inequality swaps over!)

Regarding adding the negative of the right side to both sides, using the addition and subtraction property simply add that term to both sides, the inequality sign is unaffected.


$$- {\rm{2 }} < {\rm{ }}\frac{{\left( {{\rm{x}} + {\rm{1}}} \right)}}{{\left( {{\rm{x - 3}}} \right)}}\,\,\,\,\, = > \,\,\,\,\, - {\rm{2 - }}\frac{{\left( {{\rm{x}} + {\rm{1}}} \right)}}{{\left( {{\rm{x - 3}}} \right)}} < {\rm{ }}\frac{{\left( {{\rm{x}} + {\rm{1}}} \right)}}{{\left( {{\rm{x - 3}}} \right)}} - \frac{{\left( {{\rm{x}} + {\rm{1}}} \right)}}{{\left( {{\rm{x - 3}}} \right)}}\,\,\,\,\, = > \,\,\,\,\, - {\rm{2 - }}\frac{{\left( {{\rm{x}} + {\rm{1}}} \right)}}{{\left( {{\rm{x - 3}}} \right)}} < {\rm{ 0}}$$

Use the multiplication and division property to multiply both sides by -1, this requires reversing the inequality sign.


$$( - 1)\left( { - {\rm{2 - }}\frac{{\left( {{\rm{x}} + {\rm{1}}} \right)}}{{\left( {{\rm{x - 3}}} \right)}}} \right) > ( - 1)\left( {\rm{0}} \right)\,\,\,\,\, = > \,\,\,\,\,2 + \frac{{\left( {{\rm{x}} + {\rm{1}}} \right)}}{{\left( {{\rm{x - 3}}} \right)}} > 0$$

Regarding solving the problem:


$$- {\rm{2 }} < {\rm{ }}\frac{{\left( {{\rm{x}} + {\rm{1}}} \right)}}{{\left( {{\rm{x - 3}}} \right)}}$$

Note $${\rm{x}} \ne {\rm{3}}$$ else the divisor would be zero.

If this were an equality the obvious first move would be to multiply both sides by (x-3) but according to the multiplication and division property you must reverse the inequality sign if (x-3) is negative, otherwise not. What to do?

Case 1. Require (x-3) > 0, which implies x > 3, i.e. x is required to be greater than 3 for this case. Since (x-3) > 0 we DO NOT reverse the inequality symbol, so:

$$- {\rm{2 }} < {\rm{ }}\frac{{\left( {{\rm{x}} + {\rm{1}}} \right)}}{{\left( {{\rm{x - 3}}} \right)}}\,\,\,\,\, = > \,\,\,\,\, - {\rm{2(x - 3) }} < {\rm{ }}(x + 1)\,\,\,\,\, = > \,\,\,\,\, - 2x + 6 < x + 1\,\,\,\,\, = > \,\,\,\,\,5 < 3x\,\,\,\,\, = > \,\,\,\,\,x > \frac{5}{3}$$

For case 1, x is required to BOTH be greater than 3 and greater then 5/3. If x is greater then 3 then both requirements are met, so:

...........Case 1 concludes x > 3

Case 2. Require (x-3) < 0, which implies x < 3, i.e. x is required to be less than 3 for this case. Since (x-3) < 0 we DO reverse the inequality symbol, so:

$$- {\rm{2 }} < {\rm{ }}\frac{{\left( {{\rm{x}} + {\rm{1}}} \right)}}{{\left( {{\rm{x - 3}}} \right)}}\,\,\,\,\, = > \,\,\,\,\, - {\rm{2(x - 3) > }}(x + 1)\,\,\,\,\, = > \,\,\,\,\, - 2x + 6 > x + 1\,\,\,\,\, = > \,\,\,\,\,5 > 3x\,\,\,\,\, = > \,\,\,\,\,x < \frac{5}{3}$$

For case 3, x is required to BOTH be less than 3 and less then 5/3. If x is less than 5/3 then both requirements are met, so:

...........Case 2 concludes x < 5/3

In toto, x < 5/3 or x > 3.

Note that it is "or" and not "and" since the variable x can contain only one of those values at a time.

Also, since x is required to be greater than 3 or less then 5/3 there is no possibility of the divisor (x-3) being zero.

 

[abel muroi]

Thank you!

 

[lookagain]

I'm surprised that no one showed this standard way:

From post #2:


$\displaystyle \dfrac{x + 1}{x - 3} \ + \ 2 \ > \ 0$


$\displaystyle \dfrac{x + 1}{x - 3} \ + \ \dfrac{2(x - 3)}{x - 3} \ > \ 0$


$\displaystyle \dfrac{x + 1 + 2x - 6}{x - 3} \ > 0$


$\displaystyle \dfrac{3x - 5}{x - 3} \ > \ 0, \ \ \ \ \ \ x \ne 3$


Get critical numbers, convenient test numbers, and place them on a number line.
Substitute the test numbers into the last inequality.


3x - 5 = 0

x = 5/3 -----> a critical number

- - -- - - - - --

x - 3 = 0

x = 3 ------> another critical number



Test numbers

I chose x = 0, 2, 4.


$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \$positive$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ negative$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$positive

< --------- 0 ---------- 5/3 ------------ 2 --------------------- 3 ----------------- 4 ------------- >



$\displaystyle (-\infty, \ 5/3) \ \cup \ (3, \ \infty)$