Problem 21 - Algebra and Trigonometry

Question

Contain linear equations with constants in denominators. Solve each equation. $$\frac{3 x}{5}=\frac{2 x}{3}+1$$

Step-by-Step Solution

Answer

The solution for the equation is \( x = -15 \).

Step 1: Clear the Fraction

Multiply both sides of the equation by the least common denominator, which is 15 in this case. This would eliminate the fraction. So, we can rewrite the equation as \(3x * 3 = 2x * 5 + 15 * 1\), which simplifies to \(9x = 10x + 15\).

Step 2: Simplify and Solve for x

Rearrange the equation to form a simple linear equation by isolating x. After subtracting \(10x\) from both sides, the equation is \( -x = 15\). Then, divide by -1 on both sides to solve for x, \( x = -15 \).

Key Concepts

Least Common Denominator

When working with linear equations that include fractions, finding the least common denominator (LCD) is a crucial first step. The LCD is the smallest number that each of the denominators can divide into without leaving a remainder.

For instance, if we look at denominators 5 and 3, the LCD is 15. That's because 15 is the smallest number that's evenly divisible by both 3 and 5. The purpose of finding the LCD is to clear fractions from the equation, making it easier to solve. By multiplying every term of the equation by the LCD, we convert an equation with fractions into one without, simplifying the solution process.

It's like finding a common ground where all terms can meet without any barriers—think of the denominators as different languages. The LCD is the 'universal translator' that helps these terms communicate smoothly.

Clearing Fractions in Equations

Clearing fractions in equations is similar to tidying up a room—everything becomes clearer and easier to manage. The technique involves multiplying every term in the equation by the least common denominator (LCD) to get rid of fractions.

To clear the fractions from \frac{3x}{5} = \frac{2x}{3} + 1\, you must first identify the LCD, which, as previously mentioned, is 15. This means we multiply each term by 15, effectively 'cleaning up' the fractions. The equation then simplifies to \(3x \cdot 3 = 2x \cdot 5 + 15 \cdot 1\), which is much easier to handle since it's free from fractions.

This technique is crucial because it paves the way for using simple arithmetic to manipulate algebraic equations, allowing you to focus solely on solving for the variable.

Isolating Variables

Isolating the variable is the algebraic equivalent of getting to the heart of the matter. The ultimate goal is to get the variable—usually \(x\) or \(y\)—by itself on one side of the equation. This makes the solution clear and unambiguous.

In our example, after clearing the fractions and arriving at \(9x = 10x + 15\), we continue to isolate \(x\) by moving all terms containing \(x\) to one side. By subtracting \(10x\) from both sides, we get \( -x = 15\). Finally, we divide each side by \( -1\), which cancels out the negative sign, and voilà, \(x\) stands alone on one side of the equation: \(x = -15\).

Mastering the skill of isolating variables is like honing a superpower that enables you to unveil the unknown with confidence and precision in algebraic equations.