# Solving Linear Equality with Modulus

1. Nov 10, 2008

### wanchosen

I have been trying to solve the following inequality.

Solve the equation:

(2x+3)/(x-5)<¦(4x+12)/(x+1)¦

(where ¦¦ is meant to be modulus)

I tried to solve this, firstly by using the positive modulus:

(2x+3)/(x-5)<(4x+12)/(x+1)

which gives me two solutions:-

1) x< (13-SQRT(673))/4
and
2) x> ((13+SQRT(673))/4

I then tried to solve

(2x+3)/(x-5)>(-4x-12)/(x+1)

from which I got the following 2 solutions:-

3) (1+3SQRT(17))/4

and

4) (1-3SQRT(17))/4

If I sketch a graph of the equations, I can see that

x < (13-SQRT(673))/4
and
x > (13+SQRT(673))/4

are solutions, and also

(1-3SQRT(17))/4 < x < 5

where there is an asymptote (x = 5) for the LHS linear equation.

My question is have I used the right method in solving this type of inequality, and if so how do I disregard my result for (1+3SQRT(17))/4 and how should I treat the asymptote at x=-1 for the RHS equation.

Thanks

2. Nov 10, 2008

### HallsofIvy

Staff Emeritus
The best way to solve a complicated inequality is to solve the related equation.'

To solve (2x+ 3)/(x- 5)< |(4x+12)/(x+1)|, look at (2x+3)/(x-5)= |(4x+5)/(x+1)|.

If (4x+5)/(x+1) is not negative, we have (2x+ 3)/(x-5)= (4x+5)/(x+1) and multiplying on both sides by (x-5)(x+1), (2x+3)(x+1)= (4x+5)(x-5) or 2x2+ 5x+ 3= 4x2- 15x- 25 which reduces to 2x2- 20x- 22= 0 or x2- 10x- 11= (x+1)(x-11)= 0. These are equal only at x= -1 and 11

The point of that is that "<" can change to ">" only where we have "=" or where the functions are not continuous- which for rational functions means the denominators are 0. Here that happens at x= -3, -3/2, -1, and 5. Those 5 numbers, -3, -3/2, -1, 5, and 11 divide the number line into 6 intervals: x< -3, -3< x< -3/2, -3/2< x< -1, -1< x< 5, 5< x< 11, and x> 11. We can choose one value of x in each interval and evaluate the inequality at that point. Determining that the equality is true (or false) for that point tells us the same is true for all points in that interval.

3. Nov 11, 2008

### wanchosen

Thanks HallsofIvy,

Your method makes sense. Just to clarify:-

"To solve (2x+ 3)/(x- 5)< |(4x+12)/(x+1)|, look at (2x+3)/(x-5)= |(4x+5)/(x+1)|."

Did you mean "look at (2x+3)/(x-5)= |(4x+12)/(x+1)|"?

I found the first two roots by equating the equations and multiplying both sides by (x-5)(x+1).

I have attached a document which plots the graphs using an online calculator. This shows three intersections. The method above shows the 1st and 3rd intersection points which are

(13-SQRT(673))/4 (approx -3.2355)
and
((13+SQRT(673))/4 (approx 9.7355)

However, using your method how do I obtain the second intersection point which equates to

(1-3SQRT(17))/4 (approx -2.8423)

as the first function dips below the second function at this point again which is between the intervals -3 and -3/2.

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