Question

Can someone give me an example of arational equation that has an extraneous solution using the following formula?

Answer 1

\[\frac{ x+a }{ ax } = \frac{ b }{ x }\]

Answer 2

@GeniousCreation

Answer 3

1s this making any sense to you?

Answer 4

Because its not making sense to me.

Answer 5

It's a little difficult.
But is making a little sense. So yeah...

Answer 6

1 know what an extraneous solution is 1 just don't know how to get one...

Answer 7

Does b have to be the extraneous solution?

Answer 8

dont think so

Answer 9

k.

Answer 10

1 have to go eat dinner but 1 should be back in a bit. Let me know if you figure anything out.

Answer 11

Perhaps, the variables have to match so:
x=5
a=1
b=6

\[\frac{ 5+1 }{ 5(1) }=\frac{ 6 }{ 5 }\]

l these are dependent on one another.
Hope this helps!

Answer 12

@bloofoffiction : I'm not entirely sure. Wait for a few mins, but if it is really needed, you can write it down.

Answer 13

back

Answer 14

1 think the x in the equation is a variable.

Answer 15

yeah it is.

Answer 16

how would you solve it?

Answer 17

I think I already solved it.

Answer 18

Your equation was:\[\frac{ x+a }{ ax } = \frac{ b }{ x }\tag{1}\]If you cross-multiply the terms you get:\[x(x+a)=abx\]Which leads to:\[x(x+a)-abx=0\]\[\therefore x(x+a-ab)=0\]\[\therefore x=0\text{ or }x+a-ab=0\]But \(x=0\) is not valid as that would imply that equation (1) would be dividing by zero.
Therefore the only valid solution is:\[x+a-ab=0\]\[\therefore x=ab-a=a(b-1)\]

Answer 19

0.o

Answer 20

I am assuming you were trying to solve for \(x\)?

Answer 21

1'm trying to find the extraneous solution

Answer 22

The extraneous solution here was \(x=0\)

Answer 23

"Part 1. Show all work to solve for x in the equation and check the solution.
Part 2. Explain how to identify the extraneous solution and what it means."

Answer 24

It is "extraneous" because it would be mean equation (1) involves dividing by zero which is not a valid operation

Answer 25

okay so the extraneous solution is zero and x is 5?

Answer 26

where did you get x=5 from?

Answer 27

@GeniousCreation

Answer 28

Did you follow my reasoning above where I deduced that:\[x=a(b-1)\]

Answer 29

You were not here when 1 was given that answer.

Answer 30

Sorry - I don't understand what you are trying to say?

Answer 31

Okay scratch that. Can you give me an equation following the specifications you've made?

Answer 32

It might help if you posted the whole of the question you were given here

Answer 33

Task 4—Solving Rational Equations
Using the equation below as a model, fill in numbers in the place of a and b to create a rational equation that has an extraneous solution.
=
Part 1. Show all work to solve for x in the equation and check the solution.
Part 2. Explain how to identify the extraneous solution and what it means.

Answer 34

\[\frac{ x+a }{ ax } = \frac{ b }{ x }\]

Answer 35

Ok, so I showed you how this leads to:\[x=a(b-1)\]and how we removed the extraneous solution \(x=0\) and the reasons for this.

We can therefore pick any values for \(a\) and \(b\) as long as it does not lead to \(x=0\). So, for example, we cannot pick \(b=1\) as this leads to \(x=0\).

So, lets pick \(a=10\) and \(b=5\) for example to get:\[x=a(b-1)=10(5-1)=40\]

Answer 36

okay

Answer 37

Putting this back into our original equation leads to:\[\frac{x+a}{ax}=\frac{b}{x}\]\[\therefore \frac{40+10}{10\times40}=\frac{5}{40}\]\[\therefore \frac{50}{400}=\frac{5}{40}\]which is valid so our solution here works.

Answer 38

awesome.

Answer 39

Please let me know if you need any further explanations - I would be more than happy to help :)

Answer 40

You are currently my favorite person. Arigato!

Answer 41

yw my friend! :)