Problem 49 - Prealgebra

Question

Solve each equation. Check your solution. $$63=9(2 y-3)$$

Step-by-Step Solution

Answer

The solution is \(y=5\).

Step 1: Distribute Multiplication

First, distribute the 9 into the terms inside the parentheses: \[63 = 9(2y - 3)\]This simplifies to: \[63 = 18y - 27\]

Step 2: Isolate the Variable Term

Add 27 to both sides of the equation to begin isolating the variable term:\[63 + 27 = 18y - 27 + 27\]This simplifies to:\[90 = 18y\]

Step 3: Solve for the Variable

Divide both sides by 18 to solve for \(y\):\[\frac{90}{18} = \frac{18y}{18}\]Thus, \(y = 5\).

Step 4: Check the Solution

Substitute \(y = 5\) back into the original equation to verify the solution:\[63 = 9(2(5) - 3)\]This simplifies to:\[63 = 9(10 - 3)\]\[63 = 9 \times 7\]\[63 = 63\]The solution checks out, so \(y = 5\) is correct.

Key Concepts

Distributive Property

The distributive property is a fundamental concept in prealgebra and mathematics in general. It helps us to simplify expressions and solve equations. This property states that multiplying a number by a group of numbers added together is the same as doing each multiplication separately and then adding the results. In algebraic terms, the distributive property is written as:

• \(a(b + c) = ab + ac\)

Looking at the exercise, we need to distribute the 9 to both terms inside the parentheses:

• \(9(2y - 3) = 18y - 27\)

This step is crucial because it transforms the equation into a simpler form, making it easier to isolate the variable in the subsequent steps. Remember, using the distributive property correctly is essential for correctly solving equations of this type.

Isolating Variable

Once we have used the distributive property, the next step in solving an equation is to isolate the variable. Isolating a variable means taking steps to get the variable by itself on one side of the equation. This is typically done through inverse operations, which help cancel out the other terms. In our equation, after distributing, we have:

• \(63 = 18y - 27\)

To isolate \(y\), we first add 27 to both sides to cancel out -27 on the right side:

• \(63 + 27 = 18y\)
• \(90 = 18y\)

Now, \(y\) is being multiplied by 18, so we divide both sides by 18 to solve for \(y\):

• \(y = \frac{90}{18}\)
• \(y = 5\)

Being thorough with each step ensures that the equation is balanced, yielding the correct solution.

Checking Solution

After finding our solution, \(y = 5\), it's important to check our work by substituting the value back into the original equation. This step helps confirm that the solution is correct and that no errors were made. The original equation is:

• \(63 = 9(2y - 3)\)

Substituting \(y = 5\) results in:

• \(63 = 9(2(5) - 3)\)
• \(63 = 9(10 - 3)\)
• \(63 = 9 \times 7\)
• \(63 = 63\)

Since both sides of the equation are equal, the solution \(y = 5\) is verified. Checking your work minimizes mistakes and increases confidence in your problem-solving abilities.

Prealgebra

Prealgebra is the foundation of algebra and is a critical step in understanding basic mathematical concepts and operations. It usually involves working with basic arithmetic involving numbers and variables to set the stage for more complex topics in algebra. Through prealgebra, students learn to:

• Understand and apply the distributive property.
• Perform operations to isolate variables.
• Check their work by substituting solutions back into original equations.

These skills are essential as they form the basis for building more advanced mathematical understanding. Mastering prealgebra can sometimes be challenging, but practice with exercises, like solving linear equations, ensures a strong mathematical foundation.