Cannot solve the inequality

[seema singh]

Find the sum of all integer solutions of the inequality:

4−x/x−5>1/1−x?

I am multiplying the LHS (left side) by x-5 both in numerator and denominator and RHS ( right side) by 1-x both in numerator and denominator. ( i did this so i do not affect the inequality as we dont know yet if x-5 or 1-x is positive or negative so i converted them to whole squares and then rearranged)
i am getting this by rearranging-
(4-x).(x-5).((1-x)^2) > (1-x).((x-5)^2)
But then it is getting very complex and messy do anyone have an easier way to solve this?

 

[pka]

Find the sum of all integer solutions of the inequality:
4−x/x−5>1/1−x?

The question is \(\displaystyle \frac{4-x}{x-5}>\frac{1}{1-x}\).
Rewrite as \(\displaystyle \frac{4-x}{x-5}-\frac{1}{1-x}>0\)
OR \(\displaystyle \frac{4-x}{x-5}+\frac{1}{x-1}>0\)
What id the common denominator?

 

[HallsofIvy]

Recall that if a> b then ac> bc if c> 0 and ac< bc if c< 0. Also recall that multiplying both sides of an inequality by a positive number retains the inequality but multiplying both sides by a negative number reverses it.

Starting from \(\displaystyle \frac{4- x}{x- 5}>\frac{1}{1- x}\), the denominator of the first fraction, x- 5, is positive if x> 5, negative if x< 5 and the denominator of the second fraction, 1- x, is positive if x< 1, negative if x> 1.

If x< 1, the first denominator is negative so \(\displaystyle 4- x< \frac{x- 5}{1- x}\) but the denominator is positive so \(\displaystyle (4- x)(1- x)< x- 5\). \(\displaystyle x^2- 5x+ 4< x- 5\). \(\displaystyle x^2- 6x+ 9= (x- 3)^2<0\). That is not true for any x so the original inequality is not true for x< 1.

If 1< x< 5, the first denominator is now positive so \(\displaystyle 4- x> \frac{x- 5}{1- x}\) and the second denominator is still positive so \(\displaystyle (4- x)(x- 5)>x- 5\). \(\displaystyle x^2- 5x+ 4> x- 5\), \(\displaystyle x^2- 6x+ 9= (x- 3)^2> 0\). That is true for all x other than 3 so the original inequality is true for \(\displaystyle 4< x 5\).

If x> 5, The first denominator is positive so \(\displaystyle 4- x> \frac{x- 5}{1- x}\) but the second denominator is negative so \(\displaystyle (4- x)(1- x)< x- 5\). \(\displaystyle x^2- 5x+ 4< x- 5\), \(\displaystyle x^2- 6x+ 9= (x- 3)^2< 0\). That is not true for any x so the original inequality is false for all x> 5.

The inequality is true if and only if \(\displaystyle 4< x< 5.\)

 

[pka]

Starting from \(\displaystyle \frac{4- x}{x- 5}>\frac{1}{1- x}\), the denominator of the first fraction, x- 5, is positive if x> 5, negative if x< 5 and the denominator of the second fraction, 1- x, is positive if x< 1, negative if x> 1.
If x< 1, the first denominator is negative so \(\displaystyle 4- x< \frac{x- 5}{1- x}\) but the denominator is positive so \(\displaystyle (4- x)(1- x)< x- 5\). \(\displaystyle x^2- 5x+ 4< x- 5\). \(\displaystyle x^2- 6x+ 9= (x- 3)^2<0\). That is not true for any x so the original inequality is not true for x< 1.
If 1< x< 5, the first denominator is now positive so \(\displaystyle 4- x> \frac{x- 5}{1- x}\) and the second denominator is still positive so \(\displaystyle (4- x)(x- 5)>x- 5\). \(\displaystyle x^2- 5x+ 4> x- 5\), \(\displaystyle x^2- 6x+ 9= (x- 3)^2> 0\). That is true for all x other than 3 so the original inequality is true for \(\displaystyle 4< x 5\).
If x> 5, The first denominator is positive so \(\displaystyle 4- x> \frac{x- 5}{1- x}\) but the second denominator is negative so \(\displaystyle (4- x)(1- x)< x- 5\). \(\displaystyle x^2- 5x+ 4< x- 5\), \(\displaystyle x^2- 6x+ 9= (x- 3)^2< 0\). That is not true for any x so the original inequality is false for all x> 5.
The inequality is true if and only if \(\displaystyle 4< x< 5.\)

I cannot spot the error in the above. However doing as I suggest, comparing all to zero, my solution agrees with this one.
I always told my students to compare to zero to avoid the hazards of cases.

 

[Dr.Peterson]

If x< 1, the first denominator is negative so \(\displaystyle 4- x< \frac{x- 5}{1- x}\) but the denominator is positive so \(\displaystyle (4- x)(1- x)< x- 5\). \(\displaystyle x^2- 5x+ 4< x- 5\). \(\displaystyle x^2- 6x+ 9= (x- 3)^2<0\). That is not true for any x so the original inequality is not true for x< 1.

If 1< x< 5, the first denominator is now positive so \(\displaystyle 4- x> \frac{x- 5}{1- x}\) and the second denominator is still positive so \(\displaystyle (4- x)(x- 5)>x- 5\). \(\displaystyle x^2- 5x+ 4> x- 5\), \(\displaystyle x^2- 6x+ 9= (x- 3)^2> 0\). That is true for all x other than 3 so the original inequality is true for \(\displaystyle 4< x 5\).

If x> 5, The first denominator is positive so \(\displaystyle 4- x> \frac{x- 5}{1- x}\) but the second denominator is negative so \(\displaystyle (4- x)(1- x)< x- 5\). \(\displaystyle x^2- 5x+ 4< x- 5\), \(\displaystyle x^2- 6x+ 9= (x- 3)^2< 0\). That is not true for any x so the original inequality is false for all x> 5.

The inequality is true if and only if \(\displaystyle 4< x< 5.\)

The middle case is full of typos; but it ends up being true that it "is true for all x other than 3", and the answer for this case should be "1 < x < 3 or 3 < x < 5". And that is the final answer.

I tend to avoid the case method when it can be avoided, but it's a valuable tool to have.

I'd probably use something like pka's approach. But the OP's method is a very nice trick, and can work just fine if we factor at appropriate places to avoid ever having an unfactored quartic polynomial:

We have

[MATH](4-x)(x-5)((1-x)^2) > (1-x)((x-5)^2)[/MATH],​

and can then subtract the RHS to get

[MATH](4-x)(x-5)((1-x)^2) - (1-x)((x-5)^2) > 0[/MATH].​

Factoring out common factors, we get

[MATH](x-5)(1-x)[(1-x)(4-x) - (x-5)] > 0[/MATH],​

so that

[MATH](x-5)(1-x)[x^2-6x+9] > 0[/MATH],​

which finally factors as

[MATH](x-5)(1-x)(x-3)^2 > 0[/MATH].​

This is true when x-3 is nonzero, and x-5 and 1-x have the same sign, so we end up again with

1 < x < 3 or 3 < x < 5.​