- #1

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- 0

Find a

3/a+1 = 4/a

(multiply a+1) 3 = (4a + 4)/a

(multiply a) 3a = 4a + 4

(-4) 3a - 4 = 4a

(-3a) -4 = a

Is this right? or have i done this wrong? its looks easy to do (basics) but i've forgotton! Let me kno cheers!

- Thread starter james_rich
- Start date

- #1

- 23

- 0

Find a

3/a+1 = 4/a

(multiply a+1) 3 = (4a + 4)/a

(multiply a) 3a = 4a + 4

(-4) 3a - 4 = 4a

(-3a) -4 = a

Is this right? or have i done this wrong? its looks easy to do (basics) but i've forgotton! Let me kno cheers!

- #2

HallsofIvy

Science Advisor

Homework Helper

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It certainly doesn't belong in "Linear and Abstract Algebra". I'm going to move it to "general math".

The only comment I have on it is that I see no reason to subtract 4 from both sides (moving the 4 over to the right). It would be quicker and easier to subtract 3a from both sides.

- #3

- 352

- 1

You start off with

[tex] \frac {3} {a} + 1 = \frac {4} {a}[/tex] now I hope you didnt get your order of operations, and you actaully meant 3/(a+1). But assuming the oder is right, I dont get how you derived your second equation.

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

First of all, as far as I know, if you divide a on the left, you multuiply it by everything on the right. But you seem to only multiply it by the numerator. And why do you add 4 and not 1? I would have it look like this.

[tex]3 = \frac {4} {a} \cdot a - 1[/tex]

Correct me if im wrong.

[tex] \frac {3} {a} + 1 = \frac {4} {a}[/tex] now I hope you didnt get your order of operations, and you actaully meant 3/(a+1). But assuming the oder is right, I dont get how you derived your second equation.

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

First of all, as far as I know, if you divide a on the left, you multuiply it by everything on the right. But you seem to only multiply it by the numerator. And why do you add 4 and not 1? I would have it look like this.

[tex]3 = \frac {4} {a} \cdot a - 1[/tex]

Correct me if im wrong.

Last edited:

- #4

- 23

- 0

eNathan said:You start off with

[tex] \frac {1} {a} + 1 = \frac {4} {a}[/tex] now I hope you didnt get your order of operations, and you actaully meant 3/(a+1). But assuming the oder is right, I dont get how you derived your second equation.

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

First of all, as far as I know, if you divide a on the left, you multuiply it by everything on the right. But you seem to only multiply it by the numerator. And why do you add 4 and not 1? I would have it look like this.

[tex]3 = \frac {4} {a} \cdot a - 1[/tex]

Correct me if im wrong.

think ur reading into this a bit too much, its a really basic question, and i have the right answer (i just found out)

- #5

- 352

- 1

[tex] \frac {3} {a} + 1 = \frac {4} {a}[/tex]

can be reduced to (erm, not reduced, I forgot the term)

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

This is correct???

- #6

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Bed for me!

- #7

- 352

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- #8

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Solve 3=(4a+4)/a. I get a=-4.

Therefore one does not reduce to the other which means your initial assumption was probably wrong, i.e. the order is not right and the origional equation was 3/(a+1)=4/a not (3/a)+1=4/a.

- #9

- 2

- 0

My turn...

3/a + 1 = 4/a

3/a - 4/a = -1

-1/a = -1

if: -1/a = -1 then: a = -1/-1 or a = 1

3/a + 1 = 4/a

3/a - 4/a = -1

-1/a = -1

if: -1/a = -1 then: a = -1/-1 or a = 1

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