Domain: Find the domain of an expression with fractions

Examples with solutions for Domain: Find the domain of an expression with fractions

Exercise #1

Identify 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. Hence 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 general - 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 #2

Identify the field of application of the following fraction:

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

Video Solution

Step-by-Step Solution

Let's examine the following expression:

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

As we know, the only restriction that applies to division is division by 0, given that no number can be divided into 0 parts. Hence 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, in other words:

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 #3

Identify 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. Given that no number can be divided into 0 parts, 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, in other words:

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 #4

Identify 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. Given that no number can be divided into 0 parts, 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, in other words:

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 #5

x+y:32x+6=4 \frac{x+y:3}{2x+6}=4

What is the field of application of the equation?

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps to find the domain:

  • Step 1: Recognize that the expression x+y:32x+6=4\frac{x+y:3}{2x+6}=4 involves a fraction. The denominator 2x+62x + 6 must not be zero, as division by zero is undefined.
  • Step 2: Set the denominator equal to zero and solve for xx to find the values that must be excluded: 2x+6=02x + 6 = 0.
  • Step 3: Solve 2x+6=02x + 6 = 0:
    • 2x+6=02x + 6 = 0
    • 2x=62x = -6
    • x=3x = -3
  • Step 4: Conclude that the domain of the function excludes x=3x = -3, meaning x3x \neq -3.

Thus, the domain of the given expression is all real numbers except x=3x = -3. This translates to:

x3 x\operatorname{\ne}-3

Answer

x3 x\operatorname{\ne}-3

Exercise #6

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. Given that no number can be divided into 0 parts, 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 #7

Identify 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. Given that no number can be divided into 0 parts, 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, in other words:

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 #8

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 #9

3x:4y+6=6 \frac{3x:4}{y+6}=6

What is the field of application of the equation?

Video Solution

Step-by-Step Solution

To determine the field of application of the equation 3x:4y+6=6\frac{3x:4}{y+6}=6, we must identify values of yy for which the equation is defined.

  • The denominator of the given expression is y+6y + 6. In order for the expression to be defined, the denominator cannot be zero.
  • This leads us to solve the equation y+6=0y + 6 = 0.
  • Solving y+6=0y + 6 = 0 gives us y=6y = -6.
  • This means y=6y = -6 would make the denominator zero, thus the expression would be undefined for this value.

Therefore, the field of application, or the domain of the equation, is all real numbers except y=6y = -6.

We must conclude that y6 y \neq -6 .

Comparing with the provided choices, the correct answer is choice 3: y6 y \neq -6 .

Answer

y6 y\operatorname{\ne}-6

Exercise #10

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

In order to solve the equation, start by removing 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)

Open the parentheses on the left side, making use of 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)

Continue to open 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)

Simplify further:

(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

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

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 as shown below:

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

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

We obtain the following:

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

In order to remove the one-half coefficient, multiply the entire equation by 2

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

Apply the square root formula, as shown below-

2±122 \frac{2±\sqrt{12}}{2}

Apply the properties of square roots in order to simplify the square root of 12:

2±232 \frac{2±2\sqrt{3}}{2} Divide both the numerator and denominator by 2 as follows:

1±3 1±\sqrt{3}

Answer

x=1±3 x=1±\sqrt{3}

Exercise #11

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

What is the domain of the above equation?

Video Solution

Answer

x≠0

Exercise #12

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

What is the domain of the exercise?

Video Solution

Answer

x≠0

Exercise #13

22(2x1)=30 22(\frac{2}{x}-1)=30

What is the domain of the equation above?

Video Solution

Answer

x≠0

Exercise #14

What is the domain of the exercise?

5x+82x6=30 \frac{5x+8}{2x-6}=30

Video Solution

Answer

x≠3

Exercise #15

Select the field of application of the following fraction:

16+82x 16+\frac{8}{2x}

Video Solution

Answer

x0 x\neq0

Exercise #16

Select the field of application of the following fraction:

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

Video Solution

Answer

All numbers except (-2)

Exercise #17

Solve the following equation:

x3+1(x+1)2=x \frac{x^3+1}{(x+1)^2}=x

Video Solution

Answer

x=12 x=\frac{1}{2}