(4) Algebra question: Can you help?

Keep working on it. Let’s try some combinations.

Try numbers around 5,6,4. I’m getting close.

If I shuffle it around to

ab + a = bc – c

Does that help you?

@iphigeneia

That’s the point at which I get stuck. I assume I have to collect like terms, but does that turn ab into 2ab and bc into b2c?

The answer one of the following:

1. c(b-1)
———-
b+1

2. c(b+1)
————
b-1

3. c(1-b)
————
b+1

(the lines are supposed represent the line in a fraction, sorry)

the answer is*

Let me phrase it like this: Pretend you got the ab+a part by expanding brackets. How did you get there? Then there’s one more step and you’ve got your answer.

EDIT: I just looked at your last algebra question. You know all about factorising :)

1(ab+c)=1(bc-c)?

I just want to add that I don’t have a teacher, I’m doing these at work to stop my mind from rotting.

wait the second bracket could be c(b-1)

That’s cool, I enjoyed having an excuse to look at maths again.

Okay, the common factor in ab + a is a. It then becomes a(b+1). So:

a(b+1)=c(b-1)

ab + c = bc – a
ab + a = bc – c
a(b + 1) = c(b – 1)
a = c * (b+1) / (b-1)

^ Oops, the last line should be:
a = c * (b-1) / (b+1)