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Expert: - 11/9/2008


Question

Solving equation Quest
Solve the equation (2x+3)/(x-5) < MODULUS OF [(4x+12)/(x+1)]

I wasn't sure what to do so I cross multiplied to get.

(2x+3) (x+1) < (4x + 12) (x-5) as one equation, and another one with the negative sign, seeing as it is a modulus.

-2(x+3) (4x+12) < (x-5) (x+1). I then multiplied out both of these equations, and then used the quadratic formula to obtain the answers, -9.73356,-3.17786,-1.39357, amd 9.73356.

I haven't come across a problem like this before, so thank you for your help!! I have attached an image, incase there are any difficulties in reading the question.

Answer

Inequality
Questioner:  jenny
Private: no
Subject:   

 
Question:  

Solve the equation (2x+3)/(x-5) < MODULUS OF [(4x+12)/(x+1)]

I wasn't sure what to do so I cross multiplied to get.

(2x+3) (x+1) < (4x + 12) (x-5) as one equation, and another one with the negative sign, seeing as it is a modulus.

-2(x+3) (4x+12) < (x-5) (x+1). I then multiplied out both of these equations, and then used the quadratic formula to obtain the answers, -9.73356,-3.17786,-1.39357, amd 9.73356.

I haven't come across a problem like this before, so thank you for your help!! I have attached an image, incase there are any difficulties in reading the question.
....................
You mean:

(2x+3)/(x-5) < absolute value OF [(4x+12)/(x+1)]

OR

(2x+3)/(x-5) <  |(4x+12)/(x+1)|

Here's the clue:

abs(x) = x, when x is positive, and
        -x, (opposite of x) when x is negative.

So which is it for   |(4x+12)/(x+1)|??

It depends.  If (4x+12)/(x+1) > 0, use it, but if (4x+12)/(x+1) < 0, you must take -(4x+12)/(x+1).

AND YOU MUST DO THIS BEFORE ANY CROSS-MULTIPLYING.

Now (4x+12)/(x+1) has 'split points' at  x = -3 and x = -1.

If x < -3, then both top and bottom are (-), so the fraction is positive.
If x > -1, then both top and bottom are (+), so the fraction is positive.
If x is between them, they are opposite, so the fraction is negative.

If x < -3 or x > -1, just remove || and cross-multiply, then solve.
If x > -3 or x < -1, remove || with a (-) and cross-multiply, then solve.

BUT AT THE END, MAKE SURE THE SOLUTION IS IN THE CORRECT INTERVAL.  

If your solution WITHOUT the (-) is in between, then it is no good.
If your solution WITH the (-) is NOT in between, then it is no good.

In the end, you will have to check actual solutions.  Not so easy.

I attached a graph.  The left side is in black, the right side is red.

You want BLACK < RED.

Paul Klarreich

Expertise

I can answer questions in basic to advanced algebra (theory of equations, complex numbers), precalculus (functions, graphs, exponential, logarithmic, and trigonometric functions and identities), basic probability, and finite mathematics, including mathematical induction. I can also try (but not guarantee) to answer questions on Abstract Algebra -- groups, rings, etc. and Analysis -- sequences, limits, continuity. I won't understand specialized engineering or business jargon.

Experience

I taught at a two-year college for 25 years, including all subjects from algebra to third-semester calculus.

Education/Credentials
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