Question

Find all real values of \(x\) such that \(f(x)=0\) \(f(x)=\frac{3}{x-1}+\frac{4}{x-2}\)

Short answer

The real value of \(x\) that makes the function \(f(x)=0\) is \(x = \frac{10}{7}\).

Step-by-step solution

- Clear the Fractions

To clear the fractions, find a common denominator which is the product of \(x - 1\) and \(x - 2\). Multiply every term by this common denominator: \[(x-1)(x-2)[\frac{3}{x-1} + \frac{4}{x-2}] = (x-1)(x-2) * 0\]

- Simplify the equation

Simplify by canceling out common factors and distribute the remaining values: \[3*(x-2) + 4*(x-1) = 0\]

- Solve for \(x\)

Combine like terms on both sides of the equation and solve for \(x\): \[3x - 6 + 4x - 4 = 0 \Rightarrow 7x -10 = 0 \Rightarrow x = \frac{10}{7}.\] Hence, the real solution to the problem is \(x = \frac{10}{7}\).

Key concepts

Common Denominators in Rational Expressions

Finding a common denominator is key to simplifying and solving equations involving rational expressions, which are fractions containing polynomials. In essence, it's about looking for a common ground so that we can combine the fractions.

Let’s consider an equation like the one given in our original exercise, where we have two fractions, \( \frac{3}{x-1} \) and \( \frac{4}{x-2} \). Here, to add or subtract these expressions, we need a common basis, which is the least common denominator (LCD). The LCD for our fractions is the product of their unique denominators, \( (x-1)(x-2) \), because each denominator divides the LCD without a remainder.

It’s crucial for students to understand that when you multiply each term by the LCD, the original fractions are transformed into simpler expressions without denominators, making the equation easier to work with.

Clearing Fractions in Equations

Clearing fractions from an equation is a simplification technique that paves the way to more straightforward arithmetic. When an equation contains fractions with different denominators, it can seem daunting; however, multiplying every term by the common denominator eliminates these fractions.

In the original exercise, we multiplied the whole equation by \( (x-1)(x-2) \), which is the least common denominator. As a result, the denominators are 'cleared,' and we are left with integer or polynomial terms. It’s like giving a common 'currency' to all terms of an equation, which then allows us to combine them freely.

This is especially useful as it offers a visual simplification, leading to a standard polynomial equation we can solve easily. It's a moment of clarity for many students, showing them the power of algebraic manipulation. Always remember to distribute the common denominator to each term separately to successfully clear the fractions.

Finding Real Solutions

The crux of algebra is not just manipulation, but actually finding the real solutions that make an equation true. In the setting of rational expressions, once fractions are cleared and we have a polynomial equation, solving for \(x\) involves standard techniques like combining like terms and isolating the variable.

In the process described previously, we reduced the initial complex rational equation to a simple linear one: \(7x - 10 = 0\). The solution \( x = \frac{10}{7} \) emerges after we isolate \(x\) by adding \(10\) to both sides and then dividing by \(7\). This solution is meaningful only if it doesn’t make any denominator in the original equation equal to zero, since division by zero is undefined. Therefore, always check for any values that must be excluded from the set of real numbers due to constraints in the original equation.

Finding real solutions allows students to confirm that their algebraic manipulations yield practical results, which can then be applied back to the original context of the problem.