Question

solve for c
a(b-c)=d

I'm very confused. I would appreciate if someone could explain each step to me.

Answer 1

Hey there :)

Answer 2

So we have to get c on the LHS,

\[a(b-c)=d\]
\[ab-ac=d\]
Bringing ac to the LHS,
\[ac=ab-d\]
Dividing the equation by a to have only c in the LHS
\[c=\frac{ ab-d }{ a }\]
\[c=b-\frac{ d }{ a }\]

Answer 3

@turtwig Does that make sense?

Answer 4

@turtwig I am planning to leave once you understand this... so.. are you stuck anywhere?

Answer 5

Hi @mecharv , sorry I'm having some technical difficulties.

I don't understand what you did to bringing ac to the LHS.
If I have ab-ac=d, what makes it ac=ab-d?

Where did the negative sign in ad-ac go?

Answer 6

See carefully, I have taken ac to the RHS and d to the LHS

And then I have switched LHS and RHS for convinience... So negative ac becomes positive when it goes to RHS and the positive d gets a negative sign... Is this clear?

Answer 7

ab−ac=d

Bringing ac to the LHS,
ac=ab−d

It is a little confusing, I agree. He basically did two steps in one move....

I will show you both

ab−ac=d
-ac = d - ab ab subtracted from both
ac = -d + ab multiply both sides by -1

ac = ab - d just reorganizing the RHS

Is that clearer?

Answer 8

Oh ok I think I see what you guys did. But what is the reason for multiplying each side by -1? I understand it makes -ac into a positive, but is that the only reason? @mecharv @retirEEd

Answer 9

yeah otherwise you would would to divided both sides by -a, which is doable, but a little more difficult to understand

Answer 10

@retirEEd @mecharv Thank you both very much!

Answer 11

I did very little, but your welcomed.

Answer 12

@retirEEd Woah you took a lot of pain indeed :O
@turtwig Sorry for the confusion :( My bad.

Answer 13

Don't apologize @mecharv , I'm very grateful for your clear answers to this and my other question! I think you have over-estimated my intelligence is all x)