Solving Fraction Equations with Algebraic Techniques

Solving complex fraction equations is a popular topic in algebra. This article will walk you through a specific problem, where the sum of two fractions is three times their difference, and six times the smaller fraction exceeds the larger fraction by 1 1/2. We will use algebraic techniques to find the values of the fractions.

Introduction

To solve the problem, we need to define the two fractions as x (the larger fraction) and y (the smaller fraction). The problem presents two main conditions:

The sum of the fractions is three times their difference. Six times the smaller fraction exceeds the larger fraction by 1 1/2.

Setting Up the Equations

Let's set up the equations based on the given conditions:

x y 3(x - y) 6y x 1.5 (or 6y x 3/2)

Solving the Equations

Now, let's solve these equations step by step.

Step 1: Simplify the First Equation

Starting with the first equation:

x y 3x - y

Expanding the right side:

x y 3x - 3y

Rearranging the terms:

x y 3y 3x

Simplifying:

x 4y 3x

Subtract x from both sides:

4y 2x

Dividing by 2:

2y x (Equation 1)

Step 2: Substitute into the Second Equation

Now, we substitute x from Equation 1 into the second equation:

6y x 1.5 (or 6y x 3/2)

Substituting x 2y into this equation:

6y 2y 1.5 (or 6y 2y 3/2)

Subtracting 2y from both sides:

4y 1.5 (or 4y 3/2)

Dividing both sides by 4:

y 3/8

Step 3: Find x

Using Equation 1 to find x:

x 2y 2(3/8) 6/8 3/4

Conclusion

The two fractions are:

The larger fraction x 3/4 The smaller fraction y 3/8

Therefore, the fractions are 3/4 and 3/8.

Verification

To verify the solution, we can substitute these values back into the original conditions:

3/4 3/8 3(3/4 - 3/8)
Left side: 3/4 3/8 6/8 3/8 9/8
Right side: 3(3/4 - 3/8) 3(6/8 - 3/8) 3(3/8) 9/8
Both sides are equal: 9/8 9/8 6 (3/8) 3/4 1.5
Left side: 6(3/8) 18/8 9/4 2.25
Right side: 3/4 1.5 3/4 6/4 9/4 2.25
Both sides are equal: 9/4 9/4

Both conditions are satisfied, confirming our solution.

In conclusion, the fractions are 3/4 and 3/8. This problem demonstrates the application of algebraic techniques to solve complex fraction equations effectively.