Division by a Variable | Algebra | Quantitative Aptitude

We have often come across test takers confused about division by a variable. When is it allowed, when is it not allowed? Why is it allowed in some cases and not in others? What are the constraints we need to look out for?

(I) Is division by ${x}$ allowed here: ${x^2 = 10x}$ ?

(II) Is division by $x$ allowed here: ${y = 4x}$ ?

(III) Is division by $x$ allowed here: ${x^2 < 4x}$ ?

Let’s take a detailed look at all these questions today.

The basic guidelines:

1. Division by 0 is not allowed hence you cannot divide by a variable until and unless we know that it cannot be 0.

2. In case of an inequality, when you divide by a negative number, the sign of the inequality flips. So we cannot divide by a variable until and unless we know that it cannot be 0 AND whether it is positive or negative.

Let’s look at the three questions given above and try to solve them using the guidelines given above.

(I) Is division by $x$ allowed here: $x^2 = 10x$ ?

The first thing to know here is whether $x$ can be 0.

Case 1: If nothing has been said, then $x$ can be 0 and we cannot divide by $x$ . This is how we proceed in that case:

$x^2 - 10x = 0$

$x(x - 10) = 0$

$x = 0$ or $10$

Case 2: But if the question statement tells us that $x$ is not 0, then we can divide by $x$ .

$\frac{x^2}{x} = \frac{10x}{x}$

$x = 10$

Obviously, we don’t get the second solution $x = 0$ in this case. We already know that $x$ cannot be 0.

(II) Is division by $x$ allowed here: $y = 4x$ ?

Again, it’s an equation and we need to know whether $x$ can be 0.

Case 1: If $x$ can be 0, you cannot divide by $x$ . In this case $x = 0$ and $y = 0$ is one of the infinite possible solutions.

Case 2: If $x$ cannot be 0, then we can do the following:

$\frac{y}{x} = 4$

(if it helps us solve the question)

(III) Is division by $x$ allowed here: $x^2 > -4x$ ?

Here we have an inequality. Before deciding whether we can divide by $x$ or not, we need to know not only whether $x$ can be 0 but also whether $x$ is positive or negative.

Case 1: If we know nothing about the possible values that $x$ can take, then this is how we proceed.

$x^2 + 4x > 0$