How Do You Solve the Inequality (x+3)/(x-4) < 1?

[powp]

How do I sovle the following:

Need to solve this inequality

(x+3)/(x-4) < 1

This is what I have gotten to but not sure how to state the solution or if it right

7/(x-4) <0



Solve for x

2x^(2/3)-x^(1/3)-6 = 0

Thanks

 

[Curious3141]

For the first one, what can you say about an expression like $$(x-4)^$$ Is it ever negative for any real value of x ? What happens when you multiply both sides of the inequality by it ?

For the second one, do you know how to solve a quadratic ? Try putting [itex]x^{\frac{1}{3}} = y[/tex] and see what you get.[/itex]

 

[HallsofIvy]

The simplest way to solve complicated (non-linear) inequalities is to solve the equation: The only way a continuous function can change from "< 1" to "> 1" is to pass through 1. And a rational function (fraction) is continuous every where except where the denominator is 0.
First solve $$\frac{x+3}{x-4}= 1$$ which is the same as x+3= x-4. But that's equivalent to 3= -4 and is never true. That means the only way the fraction can jump from "< 1" to "> 1" is where the denominator is 0: x= 4.
The only thing remaining is to check one value above 4 and one below: if x= 5
$$\frac{5+3}{5-4}= 8> 1$$ and if x= 0 $$\frac{0+3}{0-4}= -\frac{3}{4}< 1$$.
The inequality is true for x< 4.

As Curious3141 said, let y= x^1/3 so that x^2/3= y² and the equation becomes 2y²- y- 6= 0. Solve that for y and then find x.

 

[Curious3141]

The simplest way to solve complicated (non-linear) inequalities is to solve the equation: The only way a continuous function can change from "< 1" to "> 1" is to pass through 1. And a rational function (fraction) is continuous every where except where the denominator is 0.
First solve $$\frac{x+3}{x-4}= 1$$ which is the same as x+3= x-4. But that's equivalent to 3= -4 and is never true. That means the only way the fraction can jump from "< 1" to "> 1" is where the denominator is 0: x= 4.
The only thing remaining is to check one value above 4 and one below: if x= 5
$$\frac{5+3}{5-4}= 8> 1$$ and if x= 0 $$\frac{0+3}{0-4}= -\frac{3}{4}< 1$$.
The inequality is true for x< 4.

IMHO, the more direct and comprehensible way to solve this sort of problem is simply to multiply the inequality by a term that's always positive in such a way as to remove the fraction. It's even easier in this case when the coefficient of the unknown term is the same between the numerator and denominator, that means we're going to get a simple linear inequality at the end of it. The method should be kept simple by avoiding expanding in the intermediate stages.

Here's how I would've done it :

$$\frac{x+3}{x-4} < 1$$

Multiply LHS and RHS by {(x-4)}^2, which is positive for all real x.

$$(x+3)(x-4) < {(x-4)}^2$$

Group the terms together.

$$(x+3-x+4)(x-4) < 0$$
$$7(x-4) < 0$$
$$x < 4$$

Personally, I prefer this solution because it seems more direct. It can be used in all "ratio" form inequalities, but if the coefficients of the numerator and denominator are unequal, one would end up with a quadratic inequality. Which is still very easy to solve because it remains factorised throughout.

 

[uart]

Everyone has there own favorite method of attacking this type of problem. Personally I find it easiest to just break it into regions, in this case condisder x>4 and x<4 separately.

Region 1 : x>4
x+3 < x-4
+3 < -4
false in entire region

Region 2 : x<4
x+3 > x-4 {note the reversal of the inequality due to mult by negative}
+3 > -4
true in entire region.

"Region" 3 : x=4
function is indeterminate.

So the solution is x<4