Problem 3 - Prealgebra

Question

Solve each equation. Check your solution. $$2(a-2)=3(a-5)$$

Step-by-Step Solution

Answer

The solution is \(a = 11\).

Step 1: Expand Both Sides

First, we need to simplify both sides of the equation by applying the distributive property.For the left side: \[2(a-2) = 2a - 4\]For the right side:\[3(a-5) = 3a - 15\]So the equation becomes:\[2a - 4 = 3a - 15\]

Step 2: Isolate Variable Terms

To isolate terms with the variable on one side, subtract \(2a\) from both sides:\[2a - 4 - 2a = 3a - 15 - 2a\]This simplifies to: \[-4 = a - 15\]

Step 3: Solve for Variable

To solve for \(a\), add 15 to both sides of the equation:\[-4 + 15 = a - 15 + 15\]Simplify the equation:\[11 = a\]

Step 4: Check Solution

Substitute \(a = 11\) back into the original equation to verify the solution.Original equation:\[2(a-2) = 3(a-5)\]Substituting 11:\[2(11-2) = 3(11-5)\]Calculate both sides:\[2 \times 9 = 3 \times 6\]\[18 = 18\]Since both sides of the equation are equal, \(a = 11\) is the correct solution.

Key Concepts

Distributive Property

The distributive property is a fundamental concept in algebra that allows us to simplify expressions. When you see an equation where a number is multiplied by a binomial, like \( 2(a-2) \), you apply this property. This means you distribute, or multiply, the number outside the parentheses by each term inside.
For example, in \( 2(a-2) \), you multiply 2 by both \( a \) and \( -2 \), resulting in \( 2a - 4 \).
Similarly, for \( 3(a-5) \), you distribute the 3 to get \( 3a - 15 \).
After you apply the distributive property, the equation is in a format that's easier to work with for solving.

Variable Isolation

Once you've simplified the equation using the distributive property, the next step is to isolate the variable. This means getting the term with the variable on one side of the equation and the constant numbers on the other.
In our example, after using the distributive property, you have \( 2a - 4 = 3a - 15 \).

• First, move all terms with \( a \) to one side by subtracting \( 2a \) from both sides.
• This simplifies the equation to \( -4 = a - 15 \).
• Next, get \( a \) alone by adding 15 to both sides.

This results in \( a = 11 \).
Variable isolation is crucial because it helps find the value of unknowns in equations.

Checking Solutions

After finding a potential solution, it’s essential to check if it truly satisfies the original equation. This ensures that no errors were made in the calculations.

• Return to the original problem: \( 2(a-2) = 3(a-5) \).
• Substitute the found value, \( a = 11 \), back into the equation.
• The left side becomes \( 2(11-2) = 2 \times 9 \), which equals 18.
• The right side becomes \( 3(11-5) = 3 \times 6 \), which also equals 18.

Since both sides match, we confirm our solution is correct.
Checking your work by substituting the solution back into the equation is always a good practice, helping to catch any mistakes made during solving.