# B Algebra question -- Solving 2 simultaneous equations...

1. Apr 17, 2016

### Biker

It is a bit of a long question about series, But what is important is this...

a+ 2c = 3b

(3b-2)^2 = 2a * (4c-2)

It is asking for the ratio c to a

It have tried a lot of ways. I always end up with this. (Note maybe I not noticing a mistake, Can just anyone confirm that this is solvable? )

a^2 - 4 a c + 4 = 8 c - 4 c^2

2. Apr 17, 2016

### Math_QED

You can solve this for a! I got the same expression as you. You can write it as a^2 - 4ac + 4c^2 = 8c - 4.
Now look at the a^2 - 4ac + 4c^2. You should notice something. Solve for a then and you can express c/a

3. Apr 17, 2016

### Biker

a = 2 ( +or- sqrt( c-1) + c)

Which doesn't make sense in my view, Because the answer is 5 : 2 ( c to a). Did you get the value for c:a as a number or with variables?

Here is the original question:
2a, 3b , 4c is a Arithmetic progression
2a, 3b-2, 4c-2 is a geometric sequence

Find c : a

4. Apr 17, 2016

### Math_QED

It might have been useful to include the original question. I don't have time right now. I will look at it tomorrow.

5. Apr 17, 2016

### mfig

This must involve some fancy maneuvers, if it is possible at all. I was able to reduce the single equation we get after eliminating $b$ to the forms:

$(a-2c)^2 = 8c-4$

or

$(2c-2)^2 = 4ac-a^2$

I don't see any way to get $\frac{c}{a}$ out of either of these.

6. Apr 17, 2016

### mfig

I wonder if you have left some information out of the problem statement because I get two solutions by brute force search:

$[a,b,c] = [1/4,1/2,5/8]$
$[a,b,c] = [1,2,5/2]$

Both of these give $c/a = 5/2$ and both sets fit the given progressions. Additionally, these sets fit the progressions but yields the wrong $c/a$ ratio:

$[a,b,c] = [4,2,1]$
$[a,b,c] = [16,14,13]$

These observations lead me to think there is more to the story than was given here. Is there something missing?

7. Apr 18, 2016

### Biker

Apparently, When the teacher gave me the question the he missed some variables in the geometric sequences. Sorry for inconvenience