Problem 46
Question
Find each number. Six more than the quotient of a number and 6 is \(-3 .\) Solve \(\frac{n}{6}+6=-3\)
Step-by-Step Solution
Verified Answer
The number is \(-54\).
1Step 1: Understand the Equation
You are given the equation \( \frac{n}{6} + 6 = -3 \). This equation describes a situation where six more than the quotient of a number \(n\) and 6 equals \(-3\). "Quotient" here means "division," so \( \frac{n}{6} \) represents the division of \(n\) by 6.
2Step 2: Isolate the Fraction Term
Subtract 6 from both sides of the equation to isolate the fraction term on one side. This gives us: \( \frac{n}{6} + 6 - 6 = -3 - 6 \). Simplified, this becomes \( \frac{n}{6} = -9 \).
3Step 3: Solve for the Unknown Variable
To eliminate the fraction, multiply both sides of the equation by 6 to solve for \(n\). This results in: \( n = -9 \times 6 \), which simplifies to \( n = -54 \).
4Step 4: Verify the Solution
Substitute \(n = -54\) back into the original equation to ensure it holds true: \( \frac{-54}{6} + 6 \). Simplifying, \( \frac{-54}{6} = -9 \) and \( -9 + 6 = -3 \), which matches the initial condition. Therefore, the solution is verified.
Key Concepts
Solving EquationsPrealgebra ConceptsVariable Isolation
Solving Equations
Solving an equation is like solving a puzzle. It begins with an expression that has an unknown element, usually referred to as a variable. In this exercise, the equation is \( \frac{n}{6} + 6 = -3 \). The goal is to find the value of \( n \) that makes the equation true.
To solve equations, we perform operations on both sides to keep the equation balanced, just like weighing scales. If we add, subtract, multiply, or divide on one side, we do the same on the other.
To solve equations, we perform operations on both sides to keep the equation balanced, just like weighing scales. If we add, subtract, multiply, or divide on one side, we do the same on the other.
- Step 1: Understand the given equation and the relationship of each term.
- Step 2: Apply operations that simplify and transform the equation so that the unknown variable becomes easy to isolate.
Prealgebra Concepts
Prealgebra involves the basics of mathematics and is essential for solving equations. It becomes the foundation upon which other math skills are built. For this particular problem, understanding the terms like "quotient" and "isolate" are crucial.
The term "quotient" refers to the result of dividing one number by another. Hence, \( \frac{n}{6} \) is the operation of dividing \( n \) by 6. In prealgebra, it's also important to understand simple arithmetic and how different operations interact. For instance:
The term "quotient" refers to the result of dividing one number by another. Hence, \( \frac{n}{6} \) is the operation of dividing \( n \) by 6. In prealgebra, it's also important to understand simple arithmetic and how different operations interact. For instance:
- Subtraction cancels out addition and vice versa.
- Division is the opposite of multiplication.
Variable Isolation
Variable isolation refers to the process of moving all terms involving the variable to one side of the equation so you can easily solve for its value. In the given equation \( \frac{n}{6} + 6 = -3 \), the main task is to manipulate the equation to get \( n \) alone.
Here’s how variable isolation unfolds:
Here’s how variable isolation unfolds:
- Identify: Recognize terms with the variable. Here, it is \( \frac{n}{6} \).
- Transform: Use inverse operations to cancel out other terms. We subtract 6 from both sides to neutralize the +6.
- Solve: Once the variable term is isolated (\( \frac{n}{6} = -9 \)), multiply both sides by 6 to release \( n \) from the division and get \( n = -54 \).
Other exercises in this chapter
Problem 46
Simplify each expression. $$(t+4) 3$$
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Find an equation that is equivalent to \(-9 t=18\)
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Write an equation that describes each sequence. Then find the indicated term. Eight more than five times a number is \(78 .\) Find the number.
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A ticket to a soccer game is \(\$ 12,\) a team pennant is \(\$ 7,\) and a T-shirt is \(\$ 15 .\) Write two equivalent expressions for the total cost of a group
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