Problem 65
Question
Solve equation. Use words or set notation to identify equations that have no solution, or equations that are true for all real numbers. \(4 x+1-5 x=5-(x+4)\)
Step-by-Step Solution
Verified Answer
The given equation is true for all real numbers, meaning that it has infinitely many solutions.
1Step 1: Simplify both sides
Starting with the given equation, \(4x + 1 - 5x = 5 - (x + 4)\), simplify both sides of the equality by making use of the commutative and associative properties of addition. The left side simplifies to \(-x + 1\), and the right side simplifies to \(5 - x - 4\). This results in the equation \(-x + 1 = 5 - x - 4\).
2Step 2: Collect similar terms
The next step is to collect similar terms. The equation becomes \(-x + 1 = -x + 1\). This resulted after simplifying \(5 - x - 4\) on the right side.
3Step 3: Compare both sides
Subtract \(-x\) and 1 from both sides of the equation to maintain equality. After doing this, we get \(0 = 0\).
4Step 4: Interpret the equation
The equation \(0 = 0\) is always true, for every real number. This means there are infinitely many solutions, and every real number is a solution to the initial equation. This situation occurs when the two expressions on either side of the equation are equivalent for all values of \(x\).
Key Concepts
Equation Solving StepsProperties of EqualityNo Solution EquationsIdentities in Algebra
Equation Solving Steps
Mastering the art of solving algebraic equations opens up a world of possibilities in mathematics and beyond. Begin by examining the given problem: simplify each side of the equation, group like terms, and use inverse operations to isolate the variable.
For the given exercise, we first simplified each side, which is our first step. This involves combining terms and removing parentheses. In the next step, we grouped like terms together to further streamline the equation. Finally, we worked to isolate the variable. However, in this case, the variable was eliminated, leading us to an interesting conclusion which is explained in the subsequent sections.
For the given exercise, we first simplified each side, which is our first step. This involves combining terms and removing parentheses. In the next step, we grouped like terms together to further streamline the equation. Finally, we worked to isolate the variable. However, in this case, the variable was eliminated, leading us to an interesting conclusion which is explained in the subsequent sections.
Properties of Equality
The properties of equality are fundamental tools that allow us to manipulate equations confidently. These properties include the reflexive property (\(a = a\)), symmetric property (\(a = b\rightarrow b = a\)), transitive property (\(a = b\text{ and } b = c\rightarrow a = c\)), addition and subtraction properties (\(a = b\rightarrow a \underline{\phantom{xxx}} \bSc{\text{±}} \underline{\phantom{xxx}} c = b \underline{\phantom{xxx}} \bSc{\text{±}} \underline{\phantom{xxx}} c\)), and multiplication and division properties (\(a = b\text{ and }ceq0\rightarrow a \underline{\phantom{xxx}} \bSc{\text{×}} \underline{\phantom{xxx}} (c) = b \underline{\phantom{xxx}} \bSc{\text{×}} \underline{\phantom{xxx}} (c)\)).
It's important to apply these properties with precision to ensure each algebraic operation we perform is justified. In our exercise, we primarily used the addition property to subtract terms from both sides, maintaining the balance of the equation—fundamental to the concept of equality.
It's important to apply these properties with precision to ensure each algebraic operation we perform is justified. In our exercise, we primarily used the addition property to subtract terms from both sides, maintaining the balance of the equation—fundamental to the concept of equality.
No Solution Equations
There are scenarios where an equation can be simplified to a point where it leads to a contradiction such as \(3 = 4\), which indicates that no solution exists because there is no possible value of the variable that can satisfy the equation. However, our initial equation simplifies to \(0 = 0\), which is always true, and this is not a contradiction but an identity, leading to a different conclusion called infinitely many solutions, or an identity equation.
Identities in Algebra
Identities are special equations that are true for all values of the variables they contain. In this case, once simplified, our exercise resulted in \(0 = 0\), which is an identity since it holds true regardless of the value of \(x\).
Identifying an identity is crucial because it tells you that the original equation is true for all real numbers. It is a fundamental concept that has a vast number of applications in algebra, and understanding how to recognize an identity can significantly enhance problem-solving abilities.
Identifying an identity is crucial because it tells you that the original equation is true for all real numbers. It is a fundamental concept that has a vast number of applications in algebra, and understanding how to recognize an identity can significantly enhance problem-solving abilities.
Other exercises in this chapter
Problem 65
Use both the addition and multiplication properties of inequality to solve each inequality and graph the solution set on a number line. $$5-x \leq 1$$
View solution Problem 65
Use the given information to write an equation. Let \(x\) represent the number described in each exercise. Then solve the equation and find the number. Negative
View solution Problem 65
Suppose that the local sales tax rate is \(6 \%\) and you buy a car for \(\$ 16,800\) a. How much tax is due? b. What is the car's total cost?
View solution Problem 66
Will help you prepare for the material covered in the next section. Use \(A=\frac{1}{2} b h\) to find \(h\) if \(A=30\) and \(b=12\)
View solution