Chapter 1: Problem 64
Determine which numbers in each set are solutions to the corresponding equations. $$ 6(n+2)=23 ;\\{1,3,5\\} $$
Short Answer
Expert verified
No numbers in the set \{1, 3, 5\} are solutions to the equation.
Step by step solution
01
Substitute Each Number
We have the equation \(6(n+2) = 23\) and the set \{1, 3, 5\}. To find which numbers are solutions, we need to substitute each number from the set into the equation and check if it makes the equation true. Let's start with \(n = 1\).
02
Check \(n = 1\)
Substitute \(n = 1\) into the equation: \[ 6(1+2) = 23 \]Simplify the expression inside the brackets: \[ 6 \times 3 = 18 \]Since 18 does not equal 23, \(n = 1\) is not a solution.
03
Check \(n = 3\)
Substitute \(n = 3\) into the equation: \[ 6(3+2) = 23 \]Simplify the expression inside the brackets: \[ 6 \times 5 = 30 \]Since 30 does not equal 23, \(n = 3\) is not a solution.
04
Check \(n = 5\)
Substitute \(n = 5\) into the equation: \[ 6(5+2) = 23 \]Simplify the expression inside the brackets: \[ 6 \times 7 = 42 \]Since 42 does not equal 23, \(n = 5\) is not a solution.
05
Conclusion
None of the numbers from the set \{1, 3, 5\} satisfied the equation \(6(n+2) = 23\). Therefore, there are no solutions from this set.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution method
The substitution method is a fundamental technique used in solving algebraic equations. It involves replacing a variable with a given number from a set to determine if it satisfies the equation. In our example, we started with the equation \(6(n+2) = 23\) and the number set \{1, 3, 5\}.
We substituted each number in the set for \(n\) one by one.
The substitution method is very useful for simple equations or to check solutions from a set, as it allows you to confirm the validity of each potential solution.
We substituted each number in the set for \(n\) one by one.
- For \(n = 1\), replacing it in the equation results in \(6(1+2)\).
- For \(n = 3\), it gives \(6(3+2)\).
- For \(n = 5\), the equation becomes \(6(5+2)\).
The substitution method is very useful for simple equations or to check solutions from a set, as it allows you to confirm the validity of each potential solution.
Checking solutions
Checking solutions is about determining whether a substituted value makes the original equation true. Once you've made a substitution, the next step is to see if the left side of the equation matches the right side. Using our equation \(6(n+2) = 23\) and the number set \{1, 3, 5\}, each number was tested:
- With \(n = 1\), the equation becomes \(6 \times 3 = 18\), which does not match 23.
- For \(n = 3\), it changes to \(6 \times 5 = 30\), also not equal to 23.
- Finally, with \(n = 5\), the multiplication yields \(6 \times 7 = 42\), still unequal to 23.
Simplifying algebraic expressions
Simplifying algebraic expressions involves reducing the complexity of an equation by performing operations like addition and multiplication to make calculations easier. This is a critical skill that helps in both solving equations and verifying solutions.
In our example, for each substitution, simplifying actions were performed inside the brackets before multiplying:
In our example, for each substitution, simplifying actions were performed inside the brackets before multiplying:
- For \(n = 1\), simplify \((1+2)\) to 3, giving \(6 \times 3\).
- For \(n = 3\), simplify \((3+2)\) to 5, leading to \(6 \times 5\).
- With \(n = 5\), simplify \((5+2)\) to 7, resulting in \(6 \times 7\).