Problem 41 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the equation (if possible). $$\frac{3}{x^{2}-3 x}+\frac{4}{x}=\frac{1}{x-3}$$

Step-by-Step Solution

Answer

The solution to the equation is \(x=-\frac{3}{2}, 4\).

Step 1: Simplify the equation

Combine like terms if any. In this case, there are no like terms to be combined. So, the initial equation remains \(\frac{3}{x^{2}-3 x}+\frac{4}{x}=\frac{1}{x-3}\).

Step 2: Find the common denominator

The common denominator of the fractions is \(x(x-3)\). Thus, rewrite the equation as \(\frac{3x(x-3)}{(x^{2}-3x)x}+\frac{4(x-3)}{x(x-3)}=\frac{x(x)}{(x-3)x}\). It simplifies to \(\frac{3x^{2}-9x}{x^{2}-3x}+\frac{4x-12}{x}=\frac{x^{2}}{x-3}\).

Step 3: Clear out the denominators

Clear out the denominators by multiplying through by \(x(x-3)\). The equation becomes \(3x^{2}-9x+4x-12=x^{2}, \) which can further simplify to \(2x^{2}-5x-12=0.\)

Step 4: Factor the quadratic equation

Factor the quadratic equation to (2x+3)(x-4)=0.

Step 5: Solve for x

Set each factor equal to zero and solve for \(x\). This gives \(x=-\frac{3}{2}, 4\). Check that the obtained solutions do not invalidate denominators of the original equation. Neither solution invalidates the original equation's denominators, meaning both solutions are valid.

Key Concepts

Common Denominator

When solving rational equations, a common hurdle is dealing with multiple fractions that have different denominators. Finding a common denominator is crucial because it allows us to combine the fractions into a single expression or compare them directly.

To determine a common denominator, we look for a value that each of the individual denominators can divide into. This often involves identifying the least common multiple of the denominators. In the equation given, the denominators are \(x^2-3x\), \(x\), and \(x-3\). By analyzing the factors, we see that \(x(x-3)\) can serve as a common denominator since it incorporates all individual denominators. This shared basis for all fractions simplifies comparison and further manipulation as we move towards solving the equation.

Factoring Quadratic Equations

Factoring quadratic equations is a critical step in solving them. A quadratic equation typically looks like \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are coefficients. The process of factoring converts this standard form into a product of two binomials.

In our exercise, after clearing the denominators, we get \(2x^2-5x-12=0\). This quadratic equation factors into \(2x+3\) and \(x-4\). The reason factoring is so powerful is because it uses the Zero Product Property, which states if the product of two numbers is zero, then at least one of the numbers must be zero. Thus, by setting each binomial equal to zero, we can solve for the variable \(x\).

Clearing Denominators in Equations

Clearing denominators in equations is a method used to rid the equation of fractions, making it easier to solve. This is done by multiplying every term by the least common denominator (LCD), which, as previously mentioned, is determined by finding a value that encompasses all individual denominators.

For the given problem, we multiply every term by \(x(x-3)\) to eliminate fractions. This process must be done with care to ensure every term is multiplied correctly to avoid errors. After clearing the denominators, we are left with a simpler equation that does not contain any fractions, allowing us to use standard algebraic techniques, such as factoring, to find the solutions for \(x\). It's crucial to check that the solutions found do not make any denominator zero, which would invalidate the solution as it causes undefined terms in the initial equation.