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Chapter 1: Problem 130

Find \(b\) such that \(\frac{4 x-b}{x-5}=3\) has a solution set given by \(\varnothing\)

Short Answer

Expert verified
The value of \(b\) which makes the solution set of the given equation empty is \(b=20\).

Step by step solution

01

Set the denominator equal to zero

For the given equation, set the denominator \(x-5\) equal to zero and solve for \(x\): \[x-5=0 鈫� x=5\]
02

Solve for b

Substitute \(x=5\) back into the original equation and simplify to find the value of \(b\): \[\frac{4(5)-b}{5-5}=3 鈫� 4(5)-b=3(0) 鈫� 20-b=0 鈫� b=20\]
03

Final outcome

The value \(b=20\) makes the solution set of the equation empty, as substitution of \(x=5\) into the original equation results in division by zero, hence, no solution exists.

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Key Concepts

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

Solutions of Equations
In algebra, finding the solutions of equations is a foundational skill. A solution is a value for the variable that makes the equation true. When we solve equations, we aim to find all possible values of the variables that satisfy them. For linear equations, these solutions are often found through operations that isolate the variable on one side.
  • Simplifying expressions both sides
  • Performing arithmetic operations
  • Checking potential solutions
However, it's crucial to ensure that these operations are valid, considering special cases such as division by zero, which can lead to complications such as no solutions or special solution sets like the empty set.
Empty Set
In algebra, the term 'empty set' refers to a solution set that contains no elements. It is denoted by the symbol \(\varnothing\). An empty set implies that there are no solutions for the equation under given circumstances.
  • An impossible scenario in the equation setup can lead to an empty set.
  • It often occurs in cases where the constraints of the equation can't be satisfied.
In the context of our equation \(\frac{4x-b}{x-5} = 3\), determining the condition for an empty set involved finding a specific value for \(b\) that makes the denominator zero, thus eliminating any valid solutions.
Division by Zero
Division by zero is a mathematical operation that is undefined. This means that there is no number which you can multiply by zero to get a nonzero number. In equations, if you encounter a division by zero, it indicates a fundamental problem in the equation setup.
  • If an equation requires dividing by zero, it signifies an invalid operation.
  • This invalidity leads to no solution being possible, often leaving us with an empty set.
Thus, in our exercise, setting \( x = 5 \) makes the original expression's denominator zero, causing division by zero. Therefore, the value of \(b\) which leads to division by zero eliminates any solutions, resulting in an empty set.
Linear Equations
Linear equations are equations of the first degree, meaning they involve only linear terms of one or more variables. These equations form straight lines when plotted on a graph, hence the name 'linear.'
A typical linear equation in one variable takes the form \(ax + b = c\), where \(a\), \(b\), and \(c\) are constants.
  • Solving linear equations involves combining like terms and simplifying the equation to isolate the variable.
  • They have unique or infinite solutions, unless special conditions like division by zero arise.
In this problem, we're dealing with a linear equation transformed by dividing one expression by the variable in the denominator. Finding \(b\) ensures no valid solutions exist, shown by the attempt to equate the denominator to zero which presents a unique scenario not typically expected in basic linear equations.

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Most popular questions from this chapter

What is a quadratic equation?

Write a quadratic equation in general form whose solution set is \(\\{-3,5\\}\).

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$ (-\infty,-1] \cap[-4, \infty)=[-4,-1] $$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I obtained \(-17\) for the discriminant, so there are two imaginary irrational solutions.

Solve equation by the method of your choice. $$ \frac{x-1}{x-2}+\frac{x}{x-3}=\frac{1}{x^{2}-5 x+6} $$
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