Equation system

A

Anna55

New member
Joined
Feb 27, 2011
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24
How should you solve this equation system:
ab=1
bc=2
cd=3
de=4
ea=6

First I multiplied everything and got this:
(abcde)^2=144
Later on I squared both terms:
abcde=12
I should I continue?
Thank you in advance.
 
1) b = 1/a

Substitute into 2)

Solve for c

Substitute into 3)

etc.
 
Yes I tried that also.
This is my answer:
((4)/(3/2/1/a))(a)=6
However when I tried to solve this I got an incorrect answer. Can you please show me how to do it correctly? Thank you!
 
Anna55 said:
Yes I tried that also.
Click to expand...
Please show your steps. Also, please provide the answer you believe to be correct. Thank you! :wink:
 
I multiplied each side with the lowest common denominator and got this:
4a=6(a)(1)(2)(3)
4a=36a
a=9a

According to the key in my book a=3/2.
I believe the equation in I did in the previous post is correct since if i plug in 3/2 the answer is correct.
I should you solve the equation in the previous post?
 
Hello, Anna55!

Your preliminary work is great!

How should you solve this equation system:

. . ab .= .1 . [1]
. . bc .= .2 . [2]
. . cd .= .3 . [3]
. . de .= .4 . [4]
. . ea .= .6 . [5]

First I multiplied everything and got this: .(abcde)^2 .= .144 . ---> . abcde .= .12
Click to expand...

So you have: .(ab)(cd)(e) .= .12

We know that: .[1] ab =1 . and: .[3] cd = 3

Substitute: .(1)(3)e .= .12 . ---> . e = 4

Substitute into [4]: .d(4) = 4 . ---> . d = 1

. . and so on . . .

 
Anna55 said:
Yes I tried that also.
This is my answer:
((4)/(3/2/1/a))(a)=6
Click to expand...

How did you get that mess? You did not follow the instructions I provided. Had you, you would have found BOTH answers.

a*b = 1 ==> b = 1/a

bc=2 ==> (1/a)*c = 2 ==> c = 2a

cd=3 ==> (2a)d = 3 ==> d = 3/(2a)

de=4 ==> (3/(2a))*e = 4 ==> e = 4(2a)/3 = 8a/3

ea=6 ==> (8a/3)*a = 6 ==> 8a^2 = 18 ==> 4a^2 = 9 ==> 4a^2 - 9 = 0

You should recognize that last form and solve it easily.
 
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