Problem 17 - Algebra and Trigonometry

Question

\(9- 46\) 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 equation given is \( \frac{1}{2} y - 2 = \frac{1}{3} y \). This is a linear equation in the variable \( y \) because it can be transformed into the form \( ay + b = c \).

Step 2: Move Terms Involving y to One Side

Subtract \( \frac{1}{2} y \) from both sides to collect the terms involving \( y \) on one side: \( \frac{1}{2} y - \frac{1}{2} y - 2 = \frac{1}{3} y - \frac{1}{2} y \).

Step 3: Simplify the Expression

Simplify the left-hand side to \( -2 \) and the right-hand side using a common denominator. This gives: \( -2 = \frac{1}{3} y - \frac{3}{6} y = \frac{2}{6} y = \frac{1}{6} y \).

Step 4: Solve for y

Multiply both sides by 6 to clear the fraction: \( -12 = y \).

Step 5: Verify the Solution

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

Key Concepts

Solving Linear Equations

Solving linear equations means finding the value of a variable that makes an equation true. When working with a linear equation, the goal is to isolate the variable on one side. Linear equations have the form \( ay + b = c \), where \( y \) is the variable, and \( a \), \( b \), and \( c \) are constants. In our original exercise, the equation \( \frac{1}{2} y - 2 = \frac{1}{3} y \) is a good example of a linear equation.To solve it, you need to:

• Collect all terms involving the variable on one side.
• Simplify the expression by performing basic arithmetic operations like addition, subtraction, multiplication, or division.

In our exercise, we successfully isolated \( y \) and simplified the equation step by step to find \( y = -12 \).
Each operation helps us gradually move towards a simpler form until the solution emerges.

Transformation of Equations

Transforming equations is a vital process to simplify and solve them. This involves moving terms around and combining like terms.
Transformations do not change the equality; they only help us reveal the value of the unknown variable.In the problem \( \frac{1}{2} y - 2 = \frac{1}{3} y \), the transformation steps were:

• Subtract \( \frac{1}{2} y \) from both sides to align \( y \) terms.
• This resulted in \( -2 = \frac{1}{6} y \).

Performing such transformations involves meaningful arithmetic operations.
This systematic rearrangement is essential as it helps manage and solve the equation easily.

Checking Solutions of Equations

After finding a potential solution for a linear equation, it's always good to verify it. Checking solutions ensures that the computed value correctly satisfies the equation from the beginning.For the equation \( \frac{1}{2} y - 2 = \frac{1}{3} y \), after solving for \( y \), we obtained \( y = -12 \). To check:

• Substitute \( y = -12 \) back into the equation.
• Calculate each side separately.
• The left side becomes \( \frac{1}{2}(-12) - 2 = -6 - 2 = -8 \), while the right side simplifies to \( \frac{1}{3}(-12) = -4 \).

This validation shows if both sides match, then the solution is indeed correct.
If they don’t, you must re-evaluate your steps to find and correct any mistakes. Checking is a critical practice that conserves accuracy in algebra.