SOLUTION: [(X-4)/3]-[3/(X-4)]>(1/3)X

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Question 1012038: [(X-4)/3]-[3/(X-4)]>(1/3)X
Found 3 solutions by ValorousDawn, Theo, MathTherapy:
Answer by ValorousDawn(53)   (Show Source):
You can put this solution on YOUR website!
Move all the terms to the left side. You now have

Put all the terms over a common denominator. Your "prime" factors are 3 and (x-4) so your common denominator is 3(x-4).

Simplify and combine like terms.

Find critical points. They are -7/4 and 4. Since the leading coefficient of the entire function is negative, the end term behavior is negative. You switch signs every crossing of critical point, and as such, the only time you have a positive result is the desired interval. Since it's strictly greater than
[-7/4, 4]

Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website!
this inequality is true when x > 7/4 and x < 4.

this can also be expressed as x > 1.75 x < 4.

this can also be expressed as 1.75 < x < 4.

this can be seen easily in the following graph.

the equation is discontinuous at x = 4, and has a vertical asymptote at x = 4, as you can see on the graph.

as the graph approaches x = 4 from the left, (x-4)/3 - (3/(x-4) approaches positive infinity.

as the graph approaches x = 4 from the right, (x-4)/3 - (3/(x-4) approaches minus infinity.

at x = 7/4 the graph of (x-4)/3 - (3/(x-4) is equal in value to the graph of x/3.

the graph helps you tremendously to see what is happening.

if you don't have a good graphing software available, then get one.

it helps to visualize what is happening.

otherwise, you're working blind unless you can construct a graph manually.

the one i used is at http://www.desmos.com/calculator.

one of the ways you would solve this algebraically as follows:

start with:

subtract x/3 from both sides of the equation to get:

set the equation equal to 0 to get:

solve for x to get x = 7/4.

look at your equation to see that it has a vertical asymptote at x = 4.

you have 2 checkpoints.

you have y = 0 at x = 7/4 and you have y = undefined at x = 4.

the graph will be continuous up to x = 4 and will be continuous after x = 4.

you need to check 3 intervals.

the intervals are:

x < 7/4
x > 7/4 and < 4
x > 4

when x < 7/4, y is negative.
when x > 7/4 and less than 4, y is positive.
when x > 4, y is negative

this means the graph is positive between x = 7/4 and x = 4 but not including x = 7/4 and x = 4.

that's your solution, because:

if , then

Answer by MathTherapy(10551)   (Show Source):
You can put this solution on YOUR website!

[(X-4)/3]-[3/(X-4)]>(1/3)X

If , then
Solving the above inequality results in:
Therefore, there are 2 CRITICAL points: and , and 3 INTERVALS to test:
Testing the intervals, we find the only true interval and the solution to be: