3

I have

$$a(b - c) < d$$

I want to solve for $a > ???$. I can divide off the parentheses to get

$$a < \frac{d}{b - c}$$.

But... let's say instead I subtract both sides

$$- a(b - c) > - d$$

I can absorb the negative into the parentheses.

$$a(c - b) > - d$$

Divide off the parentheses.

$$a > - \frac{d}{(c - b)}$$

And then absorb the negative into the denominator

$$a > \frac{d}{(b - c)}$$

This is exactly the opposite of what I got the first time. Where is my mistake? Can I actually solve for $a > ???$

Blue
  • 75,673
Tiny Tim
  • 423
  • 5
    You need to consider the sign of what you are dividing by. – David Mitra Oct 14 '22 at 11:39
  • I thought I was, I tried to make sure I was only ever dividing by positive numbers? ... Or do you mean if $b - c = k$, then I need to consider the sign of k. – Tiny Tim Oct 14 '22 at 11:40
  • 1
    If $b-c$ is positive, then dividing by it won't flip the comparison; but, in that case, $c-b$ is negative, so that dividing by it will flip the comparison. – Blue Oct 14 '22 at 11:41
  • $3 < 5$ and if you multiply both sides by $2$ you get $6 < 10$. That's all well and good. If you were to instead have multiplied by $-4$ you'd have $-12 > -20$. Multiplying or dividing by a negative number flips the direction of the sign. – JMoravitz Oct 14 '22 at 11:41
  • In the event you multiply or divide by an unknown, then you need to carefully consider each possibility, for whether the unknown was positive, negative, or zero. – JMoravitz Oct 14 '22 at 11:42
  • As for finding some expression of the form $a>\square$ involving your variables, it is a one-line process to divide both sides by $(b-c)$ if it was the case that $b-c$ was negative. In the event it was positive, you won't get to $a>\square$, you'll only get to $a<\square$ (barring silly things like $a>-|a|-1$) – JMoravitz Oct 14 '22 at 11:47
  • Do people really call this "solving for $a$"? (I would have said, "Isolating $a$ on one side of the inequality" or something like that.) – JBL Oct 14 '22 at 11:51

1 Answers1

5

As mentioned in the comments, the solution depends on the sign of $b-c$. There are several cases to consider.

Solving $a(b-c)<d$ for $a$:

*) If $b-c>0$, then your first solution is correct: any $a < \dfrac{d}{b-c}$ works. This is the case since you are multiplying both sides of an inequality by a positive number ($1/(b-c)$) and this does not change an inequality.

*) If $b-c<0$, then you get $a > \dfrac{d}{b-c}$. This is the case since multiplying both sides of an inequality by a negative number (such as $1/(b-c)$) flips the inequality around.

*) If $b-c=0$ and $0<d$, then any $a$ is a solution since you have $a(b-c)=a\cdot 0 = 0 <d$.

*) If $b-c=0$ and $d \leq 0$, then there is no solution since for any $a$, $a(b-c)=a\cdot 0 =0$ which is not less than $d$.

Bill Cook
  • 29,244