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 :

1(X2)=1 \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=X2 0=X-2

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

X=2 X=2

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

Suggested Topics to Practice in Advance

  1. Variables in Algebraic Expressions
  2. Equivalent Expressions
  3. Multiplication of Algebraic Expressions
  4. Simplifying Expressions (Collecting Like Terms)
  5. The Numerical Value in Algebraic Expressions

Practice Domain

Examples with solutions for Domain

Exercise #1

Select the field of application of the following fraction:

x16 \frac{x}{16}

Video Solution

Step-by-Step Solution

Let's examine the given expression:

x16 \frac{x}{16}

As we know, the only restriction that applies to division operation is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

However in the given expression:

x16 \frac{x}{16}

the denominator is 16 and:

160 16\neq0

Therefore the fraction is well defined and thus the unknown, which is in the numerator, can take any value,

Meaning - the domain (definition range) of the given expression is:

all x

(This means that we can substitute any number for the unknown x and the expression will remain well defined),

Therefore the correct answer is answer B.

Answer

All X All~X

Exercise #2

Select the domain of the following fraction:

8+x5 \frac{8+x}{5}

Video Solution

Step-by-Step Solution

The domain depends on the denominator and we can see that there is no variable in the denominator.

Therefore, the domain is all numbers.

Answer

All numbers

Exercise #3

Select the the domain of the following fraction:

6x \frac{6}{x}

Video Solution

Step-by-Step Solution

The domain of a fraction depends on the denominator.

Since you cannot divide by zero, the denominator of a fraction cannot equal zero.

Therefore, for the fraction 6x \frac{6}{x} , the domain is "All numbers except 0," since the denominator cannot equal zero.

In other words, the domain is:

x0 x\ne0

Answer

All numbers except 0

Exercise #4

Select the field of application of the following fraction:

3x+2 \frac{3}{x+2}

Video Solution

Step-by-Step Solution

Let's examine the given expression:

3x+2 \frac{3}{x+2}

As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

3x+2 \frac{3}{x+2}

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

x+20 x+2\neq0

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

x+20x2 x+2\neq0 \\ \boxed{x\neq -2}

Therefore, the domain (definition domain) of the given expression is:

x2 x\neq -2

(This means that if we substitute for the variable x any number different from(2) (-2) the expression will remain well-defined),

Therefore, the correct answer is answer D.

Note:

In a general way - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the \neq sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every aspect to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Answer

x2 x\neq-2

Exercise #5

Select the field of application of the following fraction:

82+x \frac{8}{-2+x}

Video Solution

Step-by-Step Solution

Let's examine the given expression:

82+x \frac{8}{-2+x}

As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

82+x \frac{8}{-2+x}

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

2+x0 -2+x\neq0

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

2+x0x2 -2+x\neq0 \\ \boxed{x\neq 2}

Therefore, the domain (definition domain) of the given expression is:

x2 x\neq 2

(This means that if we substitute any number different from 2 2 for x, the expression will remain well-defined),

Therefore, the correct answer is answer C.

Note:

In a general form - solving an inequality of this form, meaning, a non-graphical, but point inequality - that uses the \neq sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Answer

x2 x\neq2

Exercise #6

Select the field of application of the following fraction:

713+x \frac{7}{13+x}

Video Solution

Step-by-Step Solution

Let's examine the given expression:

713+x \frac{7}{13+x}

As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

713+x \frac{7}{13+x}

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

13+x0 13+x\neq0

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

13+x0x13 13+x\neq0 \\ \boxed{x\neq -13}

Therefore, the domain (definition domain) of the given expression is:

x13 x\neq -13

(This means that if we substitute any number different from (13) (-13) for the variable x, the expression will remain well-defined),

Therefore, the correct answer is answer D.

Note:

In a general way - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the \neq sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Answer

x13 x\neq-13

Exercise #7

Select the field of application of the following fraction:

x+83x \frac{x+8}{3x}

Video Solution

Step-by-Step Solution

Let's examine the given expression:

x+83x \frac{x+8}{3x}

As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

x+83x \frac{x+8}{3x}

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

3x0 3x\neq0

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

3x0/:3x0 3x\neq0\hspace{6pt}\text{/}:3 \\ \boxed{x\neq 0}

Therefore, the domain (definition domain) of the given expression is:

x0 x\neq 0

(This means that if we substitute any number different from 0 0 for x, the expression will remain well-defined),

Therefore, the correct answer is answer A.

Note:

In a general form - solving an inequality of this form, meaning, a non-linear, but point inequality - which uses the \neq sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every aspect to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Answer

x0 x\neq0

Exercise #8

Select the field of application of the following fraction:

xx+3 \frac{x}{x+3}

Video Solution

Step-by-Step Solution

Let's examine the given expression:

xx+3 \frac{x}{x+3}

As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

xx+3 \frac{x}{x+3}

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

x+30 x+3\neq0

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

x+30x3 x+3\neq0 \\ \boxed{x\neq -3}

Therefore, the domain (definition domain) of the given expression is:

x3 x\neq -3

(This means that if we substitute for the variable x any number different from(3) (-3) the expression will remain well-defined),

Therefore, the correct answer is answer D.

Note:

In a general form - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the \neq sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Answer

x3 x\neq-3

Exercise #9

Select the field of application of the following fraction:

8xx \frac{-8-x}{-x}

Video Solution

Step-by-Step Solution

Let's examine the given expression:

8xx \frac{-8-x}{-x}

As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

8xx \frac{-8-x}{-x}

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

x0 -x\neq0

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

x0/:(1)x0 -x\neq0 \hspace{6pt}\text{/}:(-1) \\ \boxed{x\neq 0}

Therefore, the domain (definition domain) of the given expression is:

x0 x\neq 0

(This means that if we substitute for the variable x any number different from0 0 the expression will remain well-defined),

Therefore, the correct answer is answer A.

Note:

In a general form - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the \neq sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Answer

x0 x\neq0

Exercise #10

Find the area of domain (no need to solve)

9(4x5x)=20(3x6x+1) 9(4x-\frac{5}{x})=20(3x-\frac{6}{x+1})

Video Solution

Step-by-Step Solution

The domain of the equation is the set of domain values (of the variable in the equation) for which all algebraic expressions in the equation are well defined,

From this, of course - we exclude numbers for which arithmetic operations are not defined,

In the expression on the left side of the given equation:

9(4x5x)=20(3x6x+1) 9(4x-\frac{5}{x})=20(3x-\frac{6}{x+1}) there is multiplication between fractions whose denominators contain algebraic expressions that include the variable of the equation (which we are looking for when solving the equation),

Of course, these fractions are defined as long as the expressions in their denominators are not equal to zero (since division by zero is not possible),

Therefore, the domain of definition of the variable in the equation will be obtained from the requirement that these expressions (in the denominators of the fractions) do not equal zero, meaning:

For the fraction in parentheses in the expression on the left side we get:

x0 \boxed{ x\neq0} For the fraction in parentheses in the expression on the right side we get:

x+10 x+1\neq0 \\ Let's solve the second inequality above (in the same way as solving an equation):

x+10x1 x+1\neq0 \\ \boxed{x\neq-1}

Therefore, the correct answer is answer A.

Note:

It should be noted that the above inequality is a point inequality and not a trend inequality (meaning it negates equality: () (\neq) and does not require a trend: (<,>,\leq,\geq) ) which is solved exactly like solving an equation, this is unlike solving a trend inequality where different solution rules apply depending on the type of expressions in the inequality, for example: solving a first-degree inequality with one variable (which has only first-degree and lower algebraic expressions), is solved almost identically to solving an equation, however any division or multiplication of both sides by a negative number requires reversing the trend.

Answer

x0,x1 x≠0,x≠-1

Exercise #11

Choose the field of application of the following fraction:

8x3x+2 \frac{-8-x}{-3x+2}

Video Solution

Step-by-Step Solution

Let's examine the given expression:

8x3x+2 \frac{-8-x}{-3x+2}

As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

8x3x+2 \frac{-8-x}{-3x+2}

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

3x+20 -3x+2\neq0

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

3x+203x2/:(3)x23x23 -3x+2\neq0 \\ -3x\neq-2\hspace{6pt}\text{/}:(-3)\\ x\neq\frac{-2}{-3}\\ \boxed{x\neq \frac{2}{3}}

Therefore, the domain (definition domain) of the given expression is:

x23 x\neq \frac{2}{3}

(This means that if we substitute any number different from 23 \frac{2}{3} for x, the expression will remain well-defined),

Therefore, the correct answer is answer C.

Note:

In a general form - solving an inequality of this form, meaning, a non-graphical, but point inequality - that uses the \neq sign and not the slope signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in an identical way and all rules used to solve an equation of any type are identical for it as well.

Answer

x23 x\neq\frac{2}{3}

Exercise #12

Find the area of domain (no need to solve)

7x+5=613x \frac{7}{x+5}=\frac{6}{13x}

Video Solution

Step-by-Step Solution

The domain of the equation is the set of domain values (of the variable in the equation) for which all algebraic expressions in the equation are well defined,

From this, of course - we exclude numbers for which arithmetic operations are not defined,

In the expression on the left side of the given equation:

7x+5=613x \frac{7}{x+5}=\frac{6}{13x} there is multiplication between fractions whose denominators contain algebraic expressions that include the variable of the equation (which we are looking for when solving the equation),

Of course, these fractions are defined as long as the expression in their denominators is not equal to zero (since division by zero is not possible),

Therefore, the domain of definition of the variable in the equation will be obtained from the requirement that these expressions (in the denominators of the fractions) do not equal zero, meaning:

For the fraction in the expression on the left side we get:

x+50 x+5\neq0 \\ For the fraction in the expression on the right side we get:

13x0 13x\neq0

We will solve these inequalities (in the same way as solving an equation):

x+50x5 x+5\neq0 \\ \boxed{x\neq-5}

13x0/:13x0 13x\neq0 \hspace{8pt}\text{/:13} \\ \boxed{x\neq0}

Therefore, the correct answer is answer A.

Note:

It should be noted that the above inequality is a point inequality and not a directional inequality (meaning it negates equality: () (\neq) and does not require direction: (<,>,\leq,\geq) ) which is solved exactly like solving an equation, this is unlike solving a directional inequality where different solution rules apply depending on the type of expressions in the inequality, for example: solving a first-degree inequality with one variable (which only has first-degree algebraic expressions and below), is solved almost identically to solving an equation, however, any division or multiplication of both sides by a negative number requires reversing the direction.

Answer

x0,x5 x≠0,x≠-5

Exercise #13

Solve the following equation:

(2x+1)2x+2+(x+2)22x+1=4.5x \frac{(2x+1)^2}{x+2}+\frac{(x+2)^2}{2x+1}=4.5x

Step-by-Step Solution

To solve the equation, let's start by getting rid of the denominators.

To do this, we'll multiply the denominators:

(2x+1)2(2x+1)+(x+2)2(x+2)=4.5x(2x+1)(x+2) (2x+1)^2\cdot(2x+1)+(x+2)^2\cdot(x+2)=4.5x(2x+1)(x+2)

Let's start by opening the parentheses on the left side, mainly using the distributive property:

(4x2+4x+1)(2x+1)+(x2+4x+4)(x+2)=4.5x(2x+1)(x+2) (4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x+1)(x+2)

Let's continue by opening the parentheses on the right side of the equation:

(4x2+4x+1)(2x+1)+(x2+4x+4)(x+2)=4.5x(2x2+5x+2) (4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x^2+5x+2) Let's continue and open the parentheses on the right side of the equation:

(4x2+4x+1)(2x+1)+(x2+4x+4)(x+2)=9x3+22.5x+9x (4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=9x^3+22.5x+9x

Now let's go back and simplify the parentheses on the left side of the equation:

8x3+8x2+2x+4x2+4x+1+x3+4x2+4x+2x2+8x+8=9x3+22.5x+9x 8x^3+8x^2+2x+4x^2+4x+1+x^3+4x^2+4x+2x^2+8x+8=9x^3+22.5x+9x

Let's combine like terms:

9x3+18x2+18x+9=9x3+22.5x+9x 9x^3+18x^2+18x+9=9x^3+22.5x+9x

Notice that all terms can be divided by 9, so let's do that:

x3+2x2+2x+1=x3+2.5x+x x^3+2x^2+2x+1=x^3+2.5x+x

Let's move all numbers to one side:

x3x3+2x22.5x2+2xx+9=0 x^3-x^3+2x^2-2.5x^2+2x-x+9=0

And we get:

0.5x2x1=0 0.5x^2-x-1=0

To get rid of the one-half coefficient, let's multiply the entire equation by 2

x22x2=0 x^2-2x-2=0

Now we can use the square root formula, and we get-

2±122 \frac{2±\sqrt{12}}{2}

Let's use the properties of square roots to simplify the square root of 12:

2±232 \frac{2±2\sqrt{3}}{2} Let's divide both numerator and denominator by 2 and we get:

1±3 1±\sqrt{3}

Answer

x=1±3 x=1±\sqrt{3}

Exercise #14

2x3=4x 2x-3=\frac{4}{x}

What is the domain of the exercise?

Video Solution

Answer

x≠0

Exercise #15

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

What is the domain of the above equation?

Video Solution

Answer

x≠0

Topics learned in later sections

  1. Domain of a Function