Problem 26 - Elementary Algebra

Question

Solve each of the equations. $$\frac{x-5}{3}+2=\frac{5}{9}$$

Step-by-Step Solution

Answer

The solution is \( x = \frac{2}{3} \).

Step 1: Isolate the Fraction

To begin solving the equation \( \frac{x-5}{3} + 2 = \frac{5}{9} \), first isolate the fractional term \( \frac{x-5}{3} \) by subtracting 2 from both sides: \[ \frac{x-5}{3} = \frac{5}{9} - 2 \].

Step 2: Convert to Common Denominator

In order to subtract \( 2 \) from \( \frac{5}{9} \), express \( 2 \) as a fraction with a denominator of 9: \( 2 = \frac{18}{9} \). Now subtract: \( \frac{5}{9} - \frac{18}{9} = \frac{-13}{9} \).

Step 3: Solve for x

Now that the equation is \( \frac{x-5}{3} = \frac{-13}{9} \), eliminate the fraction by multiplying both sides by 3: \[ x - 5 = \frac{-13}{9} \times 3 \].

Step 4: Calculate the Right Side

Perform the multiplication: \( \frac{-13}{9} \times 3 = \frac{-39}{9} = -\frac{13}{3} \). Now, the equation is \( x - 5 = -\frac{13}{3} \).

Step 5: Solve for x Using Inverse Operations

Add 5 to both sides to solve for \( x \): \( x = -\frac{13}{3} + 5 \).

Step 6: Convert Integer to Fraction and Add

Convert the integer 5 to a fraction with a denominator of 3: \( 5 = \frac{15}{3} \). Then add: \( -\frac{13}{3} + \frac{15}{3} = \frac{2}{3} \).

Step 7: Final Step: Conclude the Solution

This means that the solution to the equation is \( x = \frac{2}{3} \).

Key Concepts

Solving Equations

Solving equations involves finding the value of a variable that makes an equation true. To solve an equation, you perform operations to isolate the variable. This means rearranging the equation step-by-step until you find what the variable equals.

In our example equation \( \frac{x-5}{3} + 2 = \frac{5}{9} \), the goal is to solve for \( x \). We start by isolating the equation terms involving \( x \). Subtracting \( 2 \) from both sides is the first step, ensuring \( x \) is left with fewer surrounding terms. This is critical because each operation brings us closer to uncovering \( x \)'s value.

Always remember to perform the same operation on both sides of the equation to maintain balance. This is a key principle when solving equations. It ensures the equation remains true as you simplify it. Solving equations is a sequential process, breaking down a complex equation into simpler, manageable parts.

Fraction Operations

Fraction operations can initially seem challenging, but understanding them is crucial for solving equations that include fractions. Operations with fractions involve addition, subtraction, multiplication, or division. Common denominators are vital, especially for addition and subtraction tasks.

In the exercise \( \frac{5}{9} - 2 \), it's necessary to express \( 2 \) as a fraction with the same denominator as \( \frac{5}{9} \), which is 9. Hence, \( 2 \) becomes \( \frac{18}{9} \). Once the fractions \( \frac{5}{9} \) and \( \frac{18}{9} \) share a common denominator, subtraction becomes straightforward: \( \frac{5}{9} - \frac{18}{9} = \frac{-13}{9} \).

Such operations ensure that we can handle fractions within equations smoothly, enabling us to proceed with fewer hurdles and solve the equation steadily. Mastering fraction operations is an essential skill in elementary algebra, providing a strong foundation for further learning.

Isolating Variables

Isolating variables is about safely stripping down an equation to unveil the unknown value. Our objective is to have the variable alone on one side of the equation, revealing its value. This is done through inverse operations, reversing the operations applied to the variable.

In our example where \( \frac{x-5}{3} = \frac{-13}{9} \), the variable \( x \) is still intertwined with other terms. It becomes vital to remove the fraction by multiplying both sides by 3 to release \( x - 5 \) from under the fraction's constraint. This step pushes us closer to isolating \( x \). Afterward, adding 5 compensates for the \( -5 \) initially present with \( x \).

The strategy behind isolating variables lies in systematically peeling away layers until the variable is isolated. This might be achieved via adding, subtracting, multiplying, or dividing as needed. Each action is like an unwinding maneuver, each step logically linking back to revealing the variable's true value. Have confidence in these steps to solve algebraic equations effectively.