/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 117 Why is \(\frac{6 x+12}{7 x-28}\)... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 117

Why is \(\frac{6 x+12}{7 x-28}\) undefined for \(x=4 ?\)

Short Answer

Expert verified
The function is undefined for \(x=4\) because it results in a denominator of zero, and division by zero is undefined in mathematics.

Step by step solution

01

Identify the denominator

In this step, it's necessary to find the denominator of the fraction which is \(7x - 28\).
02

Set the denominator equal to zero

The denominator of the function is the expression \(7x - 28\). To find the value of x that makes the function undefined, set the denominator equal to zero, like so: \(7x - 28 = 0\).
03

Solve for x

To solve for x, add 28 to both sides of the equation and then divide by 7. The result is \(x = 4\).
04

Validate the result

Replacing x with 4 in the denominator, we get \(7*4 - 28 = 0\) which confirms that the function is undefined where the denominator equals zero. Thus, \(x=4\) makes the function undefined as dividing by zero is undefined in mathematics.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Undefined expressions
An expression is considered undefined when it does not yield a meaning or value in mathematics. Specifically, when dealing with rational expressions, an algebraic fraction like \(\frac{a}{b}\), the expression becomes undefined if the denominator \(b\) equals zero. This is because dividing any number by zero does not result in a finite number or even an infinite one: it simply has no mathematical meaning. In the given exercise, the expression \(\frac{6x+12}{7x-28}\) becomes undefined for \(x=4\) since substituting this value into the denominator \(7x-28\) calculates as zero, making the expression division by zero, leading to an undefined state.
Division by zero
Division by zero is a crucial concept in understanding undefined expressions. When a number is divided by zero, the operation defies mathematical rules, thus making the expression impossible to evaluate. Here's why:
  • Mathematically, division is the inverse of multiplication. So, if \(\frac{a}{b}\) is defined, there should be some number \(c\) that satisfies \(b \times c = a\).
  • However, if \(b=0\), no number \(c\) can resolve the equation to produce \(a\). This absence of a suitable value means it鈥檚 impossible to divide by zero.
In the solution presented, division by zero is exemplified by the setting of \(7x - 28 = 0\). Solving this equation for \(x=4\) confirms a zero denominator, thus making the original expression undefined.
Solving equations
In solving equations, especially to identify undefined expressions, the primary goal is to determine the values of the variable that satisfy the equation. For the expression \(\frac{6x+12}{7x-28}\), you must focus on the denominator because if it becomes zero, the expression becomes undefined. The key steps in solving are:
  • Isolate the denominator equation: \(7x - 28 = 0\).
  • Add 28 to both sides, resulting in \(7x = 28\).
  • Divide by 7, yielding \(x = 4\).
This systematic approach confirms that substituting \(x = 4\) into \(7x - 28\) would indeed make it zero, rendering the overall expression undefined. Therefore, solving equations helps to pinpoint the precise values that lead to mathematical impossibilities in expressions.

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Most popular questions from this chapter

Factor: \(9 x^{2}-16 .\) (Section 6.4, Example 1)

In Exercises \(130-133,\) use the \([\mathrm{GRAPH}]\) or \([\mathrm{TABLE}]\) feature of a graphing utility to determine if the polynomial on the left side of each equation has been correctly factored. If the graphs of \(y_{1}\) and \(y_{2}\) coincide, or if their corresponding table values are equal, this means that the polynomial on the left side has been correctly factored. If not, factor the polynomial correctly and then use your graphing utility to verify the factorization. $$x^{2}-6 x+9=(x-3)^{2}$$

Use the \(x\)-intercepts for the graph in \(a[-10,10,1]\) by \([-13,10,1]\) viewing rectangle to solve the quadratic equation. Check by substitution. Use the graph of \(y=x^{2}-2 x+1\) to solve \(x^{2}-2 x+1=0\).

In Exercises \(124-127,\) factor each polynomial. $$x^{2}-y^{2}+3 x+3 y$$

Make Sense? In Exercises \(115-118\), determine whether each statement "makes sense" or "does not make sense" and explain your reasoning. I factored \(9-25 x^{2}\) as \((3+5 x)(3-5 x)\) and then applied the commutative property to rewrite the factorization as \((5 x+3)(5 x-3)\)
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