# Simplifications That Complicate

by on December 17th, 2012

At some point during your high school years, your math teacher may have presented you with the following “proof” that 1 + 1 = 1. See if you can spot the problem.

1. Begin with the premise that
2. Multiply both sides by b to get:
3. Subtract from both sides:
4. Factor both sides:
5. Divide both sides by (b – c) to get:
6. Since b = c, let’s say that b = 1 and c = 1. When we plug these values into our equation, we get 1 + 1 = 1

If you didn’t spot the problem, I encourage you to go back and try to identify the error before reading any further. The error lies at the heart of this article.

The problem occurs at step #5, when we divide both sides of the equation by (bc). The problem is that, if b = c, then b c = 0, which means we’re dividing both sides by zero in step #5.

In most cases, it’s perfectly acceptable to divide both sides of an equation by the same value. For example, if 5x = 30, we can divide both sides by 5 to see that x = 6. On the other hand, dividing both sides by zero creates problems. To see why, consider the equation: (0)(7) = 0(5). Here, it would be incorrect to assume that, if we divide both sides by zero, we can simply cross out the zeros to get the equation 7 = 5.

While you may already know the perils of dividing by zero, it’s important to recognize that it’s sometimes possible divide by zero unwittingly, as was demonstrated in the “proof” that 1 + 1 = 1.

Consider the following Data Sufficiency question:

What is the value of x?

1)

2)

Let’s begin with statement 1. If we take the given equation and simplify it by dividing both sides by (x – 3), we’re neglecting to recognize that we may, in fact, be dividing by zero (which would be the case if x = 3).

If we did divide both sides by (x – 3), we’d get the equation 2x – 1 = x + 4, which can be quickly solved to get x = 5. Since there appears to be only one solution here, we might incorrectly conclude that statement 1 is sufficient, when it is not sufficient. In fact, in addition to the solution x = 5, we should recognize that x = 3 is another solution to the original equation.

By dividing both sides of the equation by an expression that could possibly equal zero, we inadvertently eliminated one of the two possible solutions.

## Important Takeaway

Before dividing both sides of an equation by some algebraic expression, confirm that you are not possibly dividing both sides by zero.

Now let’s examine statement 2. Here, we can simplify the given equation by dividing both sides by , but before doing so, we should first confirm that we’re not possibly dividing both sides by zero.

Upon closer inspection, we can see that  can never equal zero. We know this because  is always greater than or equal to zero, so when we add 1 to , the result is guaranteed to be positive. Since  can never equal zero, it’s perfectly acceptable to divide both sides of the equation by  to get: 2x – 1 = x + 2. This simplified equation can be solved to get x = 3.

Since statement 2 yields only one possible value for x, we can conclude that this statement is sufficient, which means the correct answer is B.

So, before dividing both sides of an equation by some algebraic expression, be sure to confirm that you’re not possibly dividing both sides by zero.

• This is something very new which I have never heard of !

I have one question here

"If we did divide both sides by (x – 3), we’d get the equation 2x – 1 = x + 4, which can be quickly solved to get x = 5. Since there appears to be only one solution here, we might incorrectly conclude that statement 1 is sufficient, when it is not sufficient. In fact, in addition to the solution x = 5, we should recognize that x = 3 is another solution to the original equation."

I understand that whenever we see the terms like (x-3) which could possible may take 0 value, is a stop sign and we need to think about it, but how should we recognize there could be another solution(x=3)?

Thanks

• Good question.

If both sides of an equation are multiples of some expression, say x+5, then we can be certain that whatever value of x makes that expression equal zero is also a solution.
So, for example, if the equation is (x+5)(something) = (x+5)(something else), then one of the solutions must be x = -5
From here, we can divide both sides by (x+5) to get: something = something else
Whether or not that new equation has additional solutions must be handled on a case-by-case basis.

Cheers,
Brent

• Crystal clear. Thank you.