Problem 17 - Precalculus Mathematics for Calculus

Question

The given equation is either linear or equivalent to a linear equation. Solve the equation. $$\frac{1}{2} y-2=\frac{1}{3} y$$

Step-by-Step Solution

Answer

The solution to the equation is \( y = 12 \).

Step 1: Identify the Equation Type

The given equation is \( \frac{1}{2} y - 2 = \frac{1}{3} y \). It is a linear equation since it can be represented in the standard form \( ay + b = cy \) after simplification.

Step 2: Eliminate Fractions

To eliminate fractions, find the least common multiple (LCM) of the denominators 2 and 3, which is 6. Multiply each term of the equation by 6 to clear the fractions: \[ 6 \left( \frac{1}{2} y \right) - 6(2) = 6 \left( \frac{1}{3} y \right) \] which simplifies to \[ 3y - 12 = 2y \].

Step 3: Isolate the Variable

Subtract \(2y\) from both sides to get the variable \(y\) on one side: \[ 3y - 2y = 12 \]. This simplifies to \[ y = 12 \].

Step 4: Verify the Solution

Substitute \(y = 12\) back into the original equation to check: \[ \frac{1}{2}(12) - 2 = \frac{1}{3}(12) \]. This simplifies to \[ 6 - 2 = 4 \] and \[ 4 = 4 \]. Since both sides are equal, \(y = 12\) is the correct solution.

Key Concepts

Solving Equations

Solving equations is a fundamental skill in algebra that involves finding the value of a variable that makes the equation true. The main objective here is to transform the given equation into a simpler form, ideally isolating the variable on one side of the equation while keeping it balanced. For the equation \( \frac{1}{2} y - 2 = \frac{1}{3} y \), we aim to perform operations that render the equation into the form \( y = c \), where \( c \) is a constant.
Moving things around in the equation involves using arithmetic operations—addition, subtraction, multiplication, and division—while always performing the same operation on both sides. This ensures that the equality is maintained. Think of it as a balanced scale; whatever you do to one side, you must do to the other!
Some helpful tips for solving equations include keeping track of what operations you've performed, double-checking your steps, and verifying your solution by substituting back into the original equation.

Fractions

Fractions can often make solving equations appear more challenging, but with the right approach, they can be tackled easily. When we have fractions in an equation like \( \frac{1}{2} y - 2 = \frac{1}{3} y \), the first step is often to eliminate the fractions.
This is done by finding the Least Common Multiple (LCM) of all the denominators in the equation. The LCM makes it possible to clear out the fractions by multiplying every term in the equation by this number.

• For our equation, the LCM of 2 and 3 is 6.
• Multiplying each term by 6 changes the equation to \( 3y - 12 = 2y \), effectively eliminating the fractions.

With the fractions gone, we have a simpler linear equation without fractions, which is easier to solve. Removing fractions ensures all numbers involved are integers, simplifying further calculations.

Variable Isolation

Variable isolation is the step where we focus on getting the variable by itself on one side of the equation. It's a crucial step for solving any algebraic equation.
In our simplified equation \( 3y - 12 = 2y \), we can isolate the variable \( y \) by ensuring it's the only term on one side of the equation.

• Start by subtracting \( 2y \) from both sides to combine the \( y \)-terms: \( 3y - 2y = 12 \).
• This further simplifies to \( y = 12 \).

By getting \( y \) on one side, we determine its value. When isolating the variable, always remember to perform operations that simplify the equation while keeping it balanced.
To ensure accuracy, substitute the found value back into the original equation to verify that both sides remain equal. This confirmatory step certifies that the isolated variable value is indeed the correct solution.