Quick linear equation question

  • Thread starter james_rich
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  • #1
james_rich
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Hey, its bin ages since i did any maths, and gotten a little rusty, am i doing this right when solving these equations?

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!
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
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First that's not a linear equation!

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
eNathan
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 don't 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 I am wrong.
 
Last edited:
  • #4
james_rich
23
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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 don't 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 I am 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
eNathan
352
1
I am reading into it too much? Ok, sure. But I just want to clarify (for my own purposes) if it is true that
[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
james_rich
23
0
Not sure, its 3am in Britain! I can barely concentrate on the clock, let alone do my math homework! i'll leave it in the capable hands of the smart people on this site, tho all looks good to me

Bed for me!
 
  • #7
eNathan
352
1
lol yea I know that feeling, being way past your usual bed-time, and somebody asks you to write a program or work out some math. :zzz: Its only 9 PM and I should goto bed already.
 
  • #8
es
70
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Solve (3/a)+1=4/a. I get a=1.
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
jnahey
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
 

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