Problem 63 - Elementary and Intermediate Algebra

Question

Solve each equation and check the result. If an equation has no solution, so indicate. $$ \frac{1}{3}+\frac{2}{x-3}=1 $$

Step-by-Step Solution

Answer

The solution is \( x = 6 \).

Step 1: Move Terms

First, we will isolate the fraction with the variable by moving \( \frac{1}{3} \) to the other side of the equation. This gives us: \[ \frac{2}{x-3} = 1 - \frac{1}{3} \] Solving the right side, we calculate: \[ 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3} \] Now, the equation is: \[ \frac{2}{x-3} = \frac{2}{3} \]

Step 2: Cross-Multiply

Now, to eliminate the fractions, we cross-multiply: \[ 2 \times 3 = 2 \times (x-3) \] This simplifies to: \[ 6 = 2x - 6 \]

Step 3: Solve for x

Next, we solve for \( x \) by isolating it on one side of the equation. Start by adding 6 to both sides: \[ 6 + 6 = 2x \] Simplifying, we get: \[ 12 = 2x \] Divide both sides by 2 to solve for \( x \): \[ x = \frac{12}{2} \] \[ x = 6 \]

Step 4: Check the Solution

We need to check the solution by substituting \( x = 6 \) back into the original equation: \[ \frac{1}{3} + \frac{2}{6-3} = 1 \] Simplify: \[ \frac{1}{3} + \frac{2}{3} = 1 \] The left-hand side simplifies to \( 1 \), which equals the right-hand side, confirming that the solution is correct.

Key Concepts

Cross-Multiplication

Cross-multiplication is a powerful technique for solving equations involving fractions. It allows us to eliminate fractions by creating a proportion between the numerators and denominators on each side of the equation. When you have an equation of the form \( \frac{a}{b} = \frac{c}{d} \), cross-multiplication involves multiplying the numerator of one fraction by the denominator of the other, and vice versa. This gives us the equation:

• \( a \times d = b \times c \)

This method simplifies the equation by removing fractions altogether. In the given exercise, after rearranging we arrived at \( \frac{2}{x-3} = \frac{2}{3} \). By cross-multiplying, we multiply 2 by 3 and the other 2 by \( x-3 \), simplifying this to \( 6 = 2x - 6 \). By solving the simpler equation, we effectively find the value of \( x \) without needing to deal with fractions any longer. This method is particularly useful when each side of an equation is a single fraction.

Fraction Operations

Fraction operations are essential for solving equations that involve fractions. These operations include addition, subtraction, multiplication, division, and simplification of fractions. Every fraction consists of a numerator (top number) and a denominator (bottom number). To add or subtract fractions, it is crucial to have a common denominator. Let's break down what we've done in our exercise:

• Subtracted fractions by finding a common denominator.
• Performed cross-multiplication to clear fractions.

For the subtraction:

• We turned \( 1 \) into \( \frac{3}{3} \) to match the denominator of \( \frac{1}{3} \), resulting in \( \frac{2}{3} \).

This process simplifies equations, making them easier to solve by turning anything with fractions into whole numbers. Mastery of these operations is critical when working with more complex equations involving multiple fractions or variables.

Substitution Method

The substitution method involves replacing a variable with a number or expression in order to solve equations. It is a technique especially handy for checking solutions. In our solved example, once we determined \( x = 6 \), we substituted \( 6 \) back into the original equation to verify the answer.

• This step helps ensure that no mistakes were made during the solving process.

By substituting \( x = 6 \) back into the equation \( \frac{1}{3} + \frac{2}{6-3} = 1 \), we verified that each side evaluated to 1. This confirmed that our solution was correct. The substitution method is not only useful for verification but also in systems of equations to replace one variable with another expression. It allows us to tackle equations with more than one variable effectively, simplifying them to eventually arrive at a solution. This technique highlights the value of understanding and substituting correctly to reach accurate solutions.