Problem 22 - BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Question

Solve the equation by using the LCD. Check your solution(s). $$\frac{2}{x-3}+\frac{1}{x}=\frac{x-1}{x-3}$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 1.5 + sqrt(0.75)\) and \(x = 1.5 - sqrt(0.75)\)

Step 1: Find the Least Common Denominator (LCD)

The LCD for the fractions in the equation is \(x(x - 3)\). This is because one term has a denominator of \(x\), one term has \(x - 3\), and the last term has \(x(x - 3)\). So the LCD must include all these factors.

Step 2: Rewrite the equation with the denominators all equal to the LCD

Rewriting so, the equation becomes: \(2x + (x - 3) = (x - 1)x. Simplify this to obtain: 2x + x - 3 = x^2 - x\). This is obtained by multiplying every term by \(x(x - 3)\).

Step 3: Simplify and solve for \(x\)

Combine like terms in the equation and simplify: nth term gives us \(x^2 - 3x + 3 = 0\). This a quadratic equation, it can be solved using quadratic formula: \(x = [-b ± sqrt(b^2 - 4ac)] / (2a)\). So, \(x = [3 ± sqrt((3)^2 - 4(1)(3))] / (2 * 1)\). Solving this expression gives us \(x = 1.5 ± sqrt(0.75)\). So the solutions would be \(x = 1.5 + sqrt(0.75)\) and \(x = 1.5 - sqrt(0.75)\).

Step 4: Check your solution

Substitute each result for \(x\) back into the original equation to validate if it is correct. If both sides of the equation balance, then the solution is correct.

Key Concepts

Least Common Denominator

To solve a rational equation like \( \frac{2}{x-3} + \frac{1}{x} = \frac{x-1}{x-3} \), the first step is to find the Least Common Denominator (LCD). The LCD is crucial because it allows us to combine fractions into a single equation that is easier to handle. In this case, examine the denominators: \(x-3\) and \(x\). To include these factors, the LCD is \(x(x-3)\). By multiplying each term in the equation by this LCD, you ensure all fractions have the same denominator, enabling you to eliminate the fractions. This simplification is the key to solving the equation efficiently.

Quadratic Equations

After rewriting the rational equation with a common denominator, you will often encounter a quadratic equation. In our given problem, the rational equation transforms into \(2x + x - 3 = x^2 - x\), which simplifies to \(x^2 - 3x + 3 = 0\). This is a quadratic equation, characterized by the form \(ax^2 + bx + c = 0\). Quadratic equations can be solved using various methods, including the quadratic formula, factoring, or completing the square. Here, the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is used to find the values of \(x\). The formula helps us calculate the roots efficiently, providing the solutions needed to solve the equation.

Simplification

Simplification is an essential step in solving equations, especially when dealing with algebraic expressions containing fractions. After multiplying each term in the equation by the LCD,

• Combine like terms
• Combine them into a standard form, such as \(ax^2 + bx + c = 0\)

In our exercise, this simplification leads us to \(x^2 - 3x + 3 = 0\). Simplifying reduces the complexity of the original equation, making it more manageable to solve. Proper simplification requires careful handling of algebraic terms and ensures accuracy. By simplifying, you make the equation straightforward for applying other solving methods like the quadratic formula.

Verification of Solutions

Once solutions are found, it's imperative to verify them to check if they work in the original equation. This step confirms that you're solving correctly and that there are no extraneous solutions. For verification, substitute each solution back into the original equation

• Check if both sides of the original equation hold true
• Ensure that no terms end up undefined, such as dividing by zero.

In our case, plug the solutions \( x = 1.5 + \sqrt{0.75} \) and \( x = 1.5 - \sqrt{0.75} \) into the original equation. Validate if they satisfy the equation \( \frac{2}{x-3} + \frac{1}{x} = \frac{x-1}{x-3} \). This final check safeguards the correctness of your solution.