- #1

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Problem: x/3 = 4/x+4

Answer: 1) First I found a common denominator.

3x+12(x/3) = 3x+12(4/x+4)

2) So I got x+4 = x+3

3) Then the answer would be x=-4, x=-3

Am I right?

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- Thread starter mlbmaniaco
- Start date

- #1

- 11

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Problem: x/3 = 4/x+4

Answer: 1) First I found a common denominator.

3x+12(x/3) = 3x+12(4/x+4)

2) So I got x+4 = x+3

3) Then the answer would be x=-4, x=-3

Am I right?

- #2

TD

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[tex]\frac{x}{3} = \frac{4}{{x + 4}}[/tex]

Now, multiply both sides with a common denominator to get rid of the denominators, so with for example [itex]3\left( {x + 4} \right)[/itex]

- #3

Hurkyl

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For example:

[tex]4/x+4 = \frac{4}{x} + 4[/tex]

but

[tex]4/(x+4) = \frac{4}{x+4}[/tex]

(3) is totally wrong -- if x+3 = x+4, then x=-4 and x=-3 are certainly not solutions.

But... (2) is also totally wrong. You skipped a bunch of steps, so I don't know what you're doing wrong. Multiplication by 3x+12 was a reasonable idea, though. Could you post your work?

- #4

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- #5

TD

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- #6

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so would i then have 3(3) = x2 = 4 ?

- #7

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Never Mind, I think I'll just give up. It is way to hard for me to understand

- #8

TD

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At the LHS, the 3 will cancel out with the denominator, as we wanted.

At the RHS, you can simplify it by multiplying it when the nominator.

[tex]\frac{x}{3} = \frac{4}{{x + 4}} \Leftrightarrow 3 \cdot \frac{x}{3} = 3 \cdot \frac{4}{{x + 4}} \Leftrightarrow x = \frac{{12}}{{x + 4}}[/tex]

Now, try losing the right denominator by multiplying both sides with

- #9

- 11

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x+4 * x = 12/ x+4 * x+4

Then I would get . . .

x(x+4) = 12(x+4)

Right?

If so, what do I do next?

- #10

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wait I made a mistake . . .

It would be x(x=4) = 12

Right...

It would be x(x=4) = 12

Right...

- #11

- 11

- 0

I mean x(x+4) = 12

Right

Right

- #12

Hurkyl

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Yes:

x/3 = 4/(x+4)

implies

x(x+4) = 12

x/3 = 4/(x+4)

implies

x(x+4) = 12

- #13

TD

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Correct!mlbmaniaco said:I mean x(x+4) = 12

Right

Now bring everything to 1 side and you have a quadratic equation.

Solve with the quadratic formula or by factoring.

- #14

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We are supposed to solve by factoring, so how do I do that?

- #15

TD

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[tex]x\left( {x + 4} \right) = 12 \Leftrightarrow x^2 + 4x - 12 = 0[/tex]

Personally, I would factor just by finding zeroes

The divisors of the constant (-12) are 'possible candidates'...

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