Problem 128 - Algebra and Trigonometry

Question

Solve equation by the method of your choice. $$ \frac{x-1}{x-2}+\frac{x}{x-3}=\frac{1}{x^{2}-5 x+6} $$

Step-by-Step Solution

Answer

There is no valid solution to the equation as both roots obtained (x=1 and x=2) are extraneous roots.

Step 1: Finding the least common denominator

Find the LCD by inspecting the factors of denominators: \(x-2\), \(x-3\), and \(x^2-5x+6\) which equals (x-2)(x-3).

Step 2: Convert the Equation to Standard Form

Multiply each term of the given equation by the LCD: \((x-2)(x-3)(\frac{x-1}{x-2}+\frac{x}{x-3})=(x-2)(x-3)\frac{1}{x^{2}-5 x+6}\). This simplifies to: x(x-3)+(x-1)(x-2)=1.

Step 3: Solve for x

Simplify and solve the equation x(x-3)+(x-1)(x-2)=1 to find the values of x. This gives us a quadratic equation of the form: \(2x^2 - 5x + 2 = 0\). By factorising the equation or using the quadratic formula, the solutions obtained are x=1 and x=2.

Step 4: Verify the Solution

Validate the obtained roots against the original equation. Substitute x=1 and x=2 into the original equation. For x=2 and x=3, an undefined term is obtained because the denominator becomes 0. Thus, these are extraneous roots and are not valid solutions.

Key Concepts

Least Common Denominator

When it comes to solving rational equations, one of the key steps is to find the Least Common Denominator (LCD). The LCD is the smallest expression that each of the denominators in the equation can be divided into without leaving a remainder.

Think of it as the 'common ground' that allows you to combine fractions without altering their values. To find the LCD, you examine each denominator's factors and then multiply those factors together, but only using each factor at most once, unless it appears multiple times in a single denominator.

In our exercise, we have the denominators \(x-2\), \(x-3\), and \(x^2-5x+6\), which factored is \(x-2\)(\(x-3\)). Therefore, the LCD for this problem is also \(x-2\)(\(x-3\)), as it is the product of the unique factors from all denominators. Once the LCD is identified, it's used to transform the equation into a standard form without fractions, simplifying the solving process.

Quadratic Equation

After finding the LCD, our rational equation's fractions are eliminated, resulting in a quadratic equation, which is an equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a \eq 0\).

Quadratic equations can be solved by factorization, completing the square, or using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In our exercise, after applying the LCD, we arrive at the quadratic equation \(2x^2 - 5x + 2 = 0\).

The solutions to this equation are the 'x' values that make the equation true. Remember, these solutions are possible 'candidates' for the final solution to the original rational equation and must be checked to ensure they don't produce undefined expressions before being accepted as valid solutions.

Extraneous Roots

In the context of solving equations, extraneous roots are solutions that arise from the solving process but are not valid solutions to the original equation. They often appear when you perform operations like multiplying by an expression that could be zero or taking a square root.

It's essential to verify your solutions by substituting them back into the original equation to check if they make sense. For instance, in our example, the solutions \(x = 1\) and \(x = 2\) look valid but when we plug them back into the original rational equation, they result in denominators of zero—making them undefined and therefore not valid. Such roots are considered 'extraneous' and must be excluded from the set of solutions.

Always be on the lookout for extraneous roots when dealing with rational equations or operations that can change the domain of the original equation.