Find the Domain of (x+8)/3x: Rational Expression Analysis

Domain Restrictions with Rational Expressions

Identify the field of application of the following fraction:

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

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:07 Let's find the domain of substitution.
00:10 The substitution domain helps us make sure we never divide by zero.
00:15 This means the denominator of the fraction cannot be zero.
00:20 We'll apply this formula in our exercise.
00:26 Our goal is to isolate the value of X to find the substitution domain.
00:32 That's the domain of substitution, and there you have the solution!

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Identify the field of application of the following fraction:

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

2

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.

3

Final Answer

x0 x\neq0

Key Points to Remember

Essential concepts to master this topic
  • Rule: Denominator cannot equal zero in any rational expression
  • Technique: Set 3x ≠ 0, then solve to get x ≠ 0
  • Check: Test x = 5: (5+8)/(3·5) = 13/15 works perfectly ✓

Common Mistakes

Avoid these frequent errors
  • Restricting the numerator instead of denominator
    Don't set x + 8 ≠ 0 to get x ≠ -8 = wrong domain! The numerator can equal zero (it just makes the fraction equal zero). Always focus only on what makes the denominator zero.

Practice Quiz

Test your knowledge with interactive questions

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

What is the domain of the above equation?

FAQ

Everything you need to know about this question

Why can't the denominator be zero but the numerator can?

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Division by zero is undefined in mathematics - it has no meaning! But when the numerator is zero, the fraction just equals zero, which is perfectly fine. Think of it like cutting a pizza: you can have zero pizza pieces, but you can't cut into zero groups.

What if I have a more complicated denominator like (x-2)(x+5)?

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Set the entire denominator equal to zero and solve! For (x-2)(x+5) = 0, you'd get two restrictions: x ≠ 2 and x ≠ 5. Each factor creates its own restriction.

Do I need to simplify the fraction first before finding the domain?

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No! Always find the domain using the original expression. If you simplify first, you might accidentally remove important restrictions and get the wrong domain.

How do I write the domain if there are multiple restrictions?

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List all restrictions separately! For example, if x ≠ 2 and x ≠ 5, write the domain as: x2,x5 x \neq 2, x \neq 5 or you can write it as two separate conditions.

What happens if I substitute a restricted value?

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You'll get division by zero, which makes your calculator show "Error" or "Undefined". That's exactly why we find these restrictions - to avoid this mathematical impossibility!

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