Identify Applications of the Fraction 7/(13+x): Domain Analysis

Domain Restrictions with Rational Functions

Identify the field of application of the following fraction:

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

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find the domain of assignment
00:03 Assignment domain exists, to ensure we don't divide by 0
00:06 Meaning the denominator in the fraction must be different from 0
00:09 We will use this formula in our exercise
00:19 We will isolate X to find the domain of assignment
00:24 This is the domain of assignment
00:27 And this is the solution to the question

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:

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

2

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.

3

Final Answer

x13 x\neq-13

Key Points to Remember

Essential concepts to master this topic
  • Rule: Denominators cannot equal zero in any fraction
  • Technique: Set denominator ≠ 0: 13 + x ≠ 0, so x ≠ -13
  • Check: Substitute x = -13: denominator becomes 13 + (-13) = 0 ✓

Common Mistakes

Avoid these frequent errors
  • Setting the numerator equal to zero instead of the denominator
    Don't set 7 = 0 to find restrictions = no solution exists! The numerator being zero just makes the whole fraction equal zero, not undefined. Always set the denominator ≠ 0 to find domain restrictions.

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 we divide by zero?

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Division by zero is undefined because you can't split something into zero parts! It would break all mathematical rules and create contradictions.

What does 'field of application' or 'domain' mean?

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The domain is all the x-values that make the expression defined and meaningful. It's basically asking: what values of x are allowed?

How do I solve 13 + x ≠ 0?

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Treat it like a regular equation! Subtract 13 from both sides: x13 x \neq -13 . The ≠ sign just means 'not equal to' instead of 'equals'.

What if the denominator had multiple terms like (x-2)(x+5)?

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Set each factor ≠ 0 separately: x20 x-2 \neq 0 AND x+50 x+5 \neq 0 , so x2 x \neq 2 and x5 x \neq -5 .

Can I test my answer by plugging in numbers?

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Yes! Try x = 0: 713+0=713 \frac{7}{13+0} = \frac{7}{13} works. Try x = -13: 713+(13)=70 \frac{7}{13+(-13)} = \frac{7}{0} undefined! ✓

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