# Transposition of terms and domain of equations of one unknown.

🏆Practice domain field

The transposition of terms involves passing the terms of an equation from one member to another. In fact, it is a group of numbers that, according to mathematical rules, are allowed to be placed in place of the unknown (or variable) within an equation. The concept of transposing terms is especially concerning equations with fractions or square roots in order to find the domain of the equation.

In certain cases we must pay attention to the transposition of terms:

1. In case of fraction the denominator cannot be equal to zero.
2. In case of square root, the radicand cannot be negative.

This means that, in such cases, it is not enough to solve the equation, but we must check if the given solution is sustainable or makes sense in the real numbers.

How do you find the domain of an equation with one unknown?
1. In equations with fractions we will find the domain by equaling the denominator to zero.
The value or values that cause the denominator to equal zero are outside the domain of the equation.

• In an equation with a root the values that cause the root to be negative are outside the domain of the equation.

Below we can see an example of how to find the domain of an equation using transpose of terms:

Example :

$\frac{1}{(X-2)}=1$

This is an equation with fraction in which the unknown appears in the denominator. The denominator cannot be zero, so the expression is not well defined:

$0=X-2$

Using transposition of terms we can clear the unknown and we obtain:

$X=2$

Therefore, the domain of the function is all real numbers except when $X=2$.

## Test yourself on domain field!

Select the field of application of the following fraction:

$$\frac{x}{16}$$

## Elaboration of the transposition of terms

In this article we will learn what is the domain of an equation (also called the set where the equation is well defined), we will see several examples that will teach us how to use it.

The domain of an equation (or solution set) is the set of all the real numbers that can be placed in the equation keeping it defined and correct according to the mathematical properties.

In this article we will only focus on different cases to examine which is the solution set that will allow the mathematical expression to remain legitimate:

### Example 1

We will soon see that this is not such a complicated subject. Let's start with a simple exercise.

Let's look at this equation:

$\frac{2}{X}=4$

As we have learned:

In mathematics you cannot divide by $0$. That is, doing so would result in an invalid expression!

Let us emphasize:

We will never write a $0$ in the denominator. We will always check that the denominator is not zero.

This is the first thing we will examine in our solution set.

Let's go back to the equation:

$\frac{2}{X}=4$

Let us note directly that the of the denominator is equal to zero when $X=0$, then the solution set is all $X$ that is not $0$.

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### Example 2

$\frac{2}{(5-X)}=10$

This is an equation with fraction in which the unknown appears in the denominator. The denominator cannot be zero, therefore it would be an expression that is not defined in the real numbers, then:

$5-X=0$

Using the transposition of terms we have that:

$X=5$

Therefore, the domain of the equation is all real numbers such that. $X≠5$

### Example 3

$\frac{4}{X-1}=2$

This is an equation with a fraction in which the unknown appears in the denominator. The denominator cannot be zero, therefore it would not be real, so:

$X-1=0$

Using transposition of terms we have that:

$X=1$

Therefore, the domain of the equation is all real numbers except when. $X=1$

Do you know what the answer is?

### Example 4

In the exercise

$\sqrt{X}=4$

We check that the expression in the root is not negative.

We write it down as follows:

or, in words: the solution set is all $X$ equal to or greater than zero.

$\sqrt{X}=4$

We will make sure that the expression inside the root is not negative.

We will write it like this:

$X≥0$

Therefore, the domain of the equation is any $X$ that is greater than or equal to zero.

### Example 4 (two different denominators)

This is a slightly more complex example, but it will help us understand the meaning of the solution set.

Even if for now you don't know how to solve such an equation that's fine, the important thing at this point is the set of real numbers where the equation is well defined.

Let's look at the equation:

$\frac{X-1}{X-2}=\frac{1}{X-3}$

Recall that the denominator cannot be $0$. In this case we have two fractions, that is, two different denominators. We must verify that neither of them is equal to zero. We will find out which is the set where the equation is well defined, by what we see when the denominators are equal to zero:

$X-2=0$

y

$X-3=0$

Using the transpose of terms we obtain that the equation is not defined with:

$X=2$

$X=3$

That is, the equation is well defined for all real numbers except for $X=2$ and $X=3$.

If we had more denominators we would have had to check each of them to make sure that none of them had the value $0$.

### Example 5

Equation without solution (exercise for advanced students)

This is an exercise for advanced students (requires understanding of the solution of equations and knowledge with equations that have no solution).

Let's look at the equation:

$\frac{X-1}{X-2}=\frac{1}{X-2}$

First, as in all cases where we have a variable in the denominator, we will find out which is the set of numbers where the equation is well defined.

We must corroborate that the denominator is not equivalent to $0$.

In this case it is clear that the solution set is:

All real numbers X with $X≠2$

Now we will continue with the exercise.

We will multiply both members of the equation by the common denominator and we will obtain:

$\frac{X-1}{X-2} \times (x-2) =\frac{1}{X-2} \times (x-2)$

$X-1=1$

$X=2$

That is, we get the result $X=2$. However, let us remember the set where the equation is well defined is:

All real numbers X such that $X≠2$

That means that the result we obtained is not within the domain of the equation. Therefore, this exercise has no solution. This exercise highlights why it is so important that we always check the domain of the equations. .

## Now we will present another type of exercise with a slightly different set of solutions.

### One-variable equations involving roots

#### Let's look at the exercise

$\sqrt{X}=4$

When we have an unknown in the root, we will remember that the expression inside the root cannot be negative, that is, it must be equal to or greater than $0$. Otherwise it would be an invalid expression.

#### For example, the expression

$\sqrt{(-5)}$

is not valid and is not mathematically defined.

The expression:$\sqrt{0}$

is valid: $\sqrt{0}=0$

Let's go back to our exercise

$\sqrt{X}=4$

First, we will find out which is the set of numbers where the equation is well defined, that is, we will check that the expression at the root is not negative. We will write it down as follows:

$X≥0$

or, in other words: the set of numbers where the equation is well-defined when $X$ is equal to or greater than zero.

Do you think you will be able to solve it?

### Example 6

Find the set of numbers where the equation is well defined.

$\sqrt{X-2}=3$

Recall that we must check that the expression at the root is not negative, so we will write:

$X−2≥0$

That is,

$X≥2$

the solution set is any real number $X$ where $X$ is equal to or greater than $2$.

There are other cases in which we could obtain invalid or undefined mathematical expressions that influence the domain of the equation, but for now it will suffice for us to check the following:

• The denominator is never $0$
• The expression at the root is not negative

## Questions on the subject

When is an equation involving a fraction not well defined?

When the denominator is equal to zero, that is, the bottom part of the fraction.

When is an equation involving a root not well defined?

When the radicand is less than zero, i.e., the number inside the root cannot be negative.

What is the transposition method?

It is the method used in equations with one unknown to isolate the variable.

How to find the domain of an equation?

With the method of transposition of terms we identify numbers where the equation is not well defined, i.e., numbers where a denominator of a fraction is zero or where the radicand of a root is negative.

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## examples with solutions for domain field

### Exercise #1

Select the field of application of the following fraction:

$\frac{x}{16}$

### Video Solution

$All~X$

### Exercise #2

Select the field of application of the following fraction:

$\frac{8+x}{5}$

All numbers

### Exercise #3

Select the field of application of the following fraction:

$\frac{6}{x}$

### Video Solution

All numbers except 0

### Exercise #4

$2x+\frac{6}{x}=18$

What is the domain of the above equation?

x≠0

### Exercise #5

$2x-3=\frac{4}{x}$

What is the domain of the exercise?