Problem 44 - Introductory Algebra for College Students

Question

Solve each equation and check your proposed solution in Exercises. Begin your work by rewriting each equation without fractions. $$\frac{3 x}{5}-\frac{2}{5}=\frac{x}{3}+\frac{2}{5}$$

Step-by-Step Solution

Answer

The solution to the equation is \(x = 3\).

Step 1: Eliminate the Fractions

To eliminate the fractions, determine the least common multiple (LCM) of the denominators which in this case is 15. Multiply each term of the equation by 15, so it becomes: \(15 \cdot \frac{3x}{5} - 15 \cdot \frac{2}{5} = 15 \cdot \frac{x}{3} + 15 \cdot \frac{2}{5}\), which can be simplified to: \(9x - 6 = 5x + 6\).

Step 2: Solve for x

Solve for \(x\) by bringing every term containing \(x\) to one side and constant terms to the other side and solve the equation. In our equation, we move \(5x\) to the left side, resulting in: \(9x-5x = 6+6\) which simplifies to: \(4x = 12\). Divide by 4 on both sides to obtain \: \(x = 3\).

Step 3: Check the Solution

To check the solution, substitute \(x\) in the original equation: \(\frac{3 \cdot 3}{5} - \frac{2}{5} = ?\) and \(\frac{3}{3} + \frac{2}{5} = ?\). Simplify and see if both sides equal. If they do, then \(x = 3\) is a valid solution.

Key Concepts

Eliminating Fractions in Equations

When you encounter equations with fractions, it can be challenging to find a solution. However, you can eliminate these fractions by finding a common denominator for all the fractional terms and then multiplying each term of the equation by this common number. This method simplifies the equation and gets rid of denominators, making it easier to solve.

For example, if you have the equation \( \frac{3x}{5} - \frac{2}{5} = \frac{x}{3} + \frac{2}{5} \), you first identify the least common multiple (LCM) of the denominators (5 and 3), which is 15. By multiplying every term by 15, you can remove the fractions and simplify the equation further to solve for the unknown variable, in this case, 'x'.

Finding Least Common Multiple

The least common multiple (LCM) of two or more numbers is the smallest multiple that is exactly divisible by each of the numbers. Finding the LCM is crucial when dealing with equations involving fractions, as it helps in converting the fractions to whole numbers.

In our equation, the LCM of 5 and 3 is 15. By determining the LCM, you multiply each term of the equation by this number to effectively eliminate the fractions, simplifying the algebraic expression and paving the way for an easier solution.

Checking Algebraic Solutions

After solving an algebraic equation, it's important to verify the solution to ensure it's correct. You can check your solution by substituting the value of the unknown variable back into the original equation and simplifying. If both sides of the equation are equal after the substitution, you have the correct solution.

Following our previous example, if you find that 'x' equals 3, substitute '3' back into the original equation \( \frac{3x}{5} - \frac{2}{5} \) on the left side and \( \frac{x}{3} + \frac{2}{5} \) on the right side. Both sides should evaluate to the same numerical value, confirming that 'x = 3' is indeed the correct solution.

Simplifying Algebraic Expressions

Simplification of algebraic expressions is a fundamental technique in algebra. It involves performing operations like addition, subtraction, multiplication, and division to reduce expressions to a simpler form. Simplifying allows you to combine like terms, reduce fractions, and make equations manageable and solvable.

In our equation example, after eliminating fractions and rearranging terms, the expression can be simplified by combining like terms, such as \(9x - 5x\), which simplifies to \(4x\). Afterwards, solving for 'x' becomes straightforward and requires simple arithmetic.