Problem 71 - Algebra 2

Question

Solve each equation. Check your answers. \(\frac{5}{x^{2}-x}+\frac{3}{x-1}=6\)

Step-by-Step Solution

Answer

The solution to the equation is \(x = \frac{5}{3}\). The other solution \(x = 1\) was found to be an extraneous solution and discarded

Step 1: Combine Fractions

The expression \(\frac{5}{x^{2}-x}\) can be simplified to \(\frac{5}{x(x-1)}\). Now, we have: \(\frac{5}{x(x-1)}+\frac{3}{x-1}=6\)

Step 2: Multiply through by the common denominator

We have a common denominator of x(x-1). To eliminate this denominator, we multiply each term in the equation by x(x-1): \(5+x(x-1)(3)=6x(x-1)\). This simplifies to: \(5+3x^{2}-3x=6x^{2}-6x\)

Step 3: Rearrange the equation and isolate terms

Subtract 3x^2 - 3x from both sides: \(3x^{2}-6x^{2}+3x-6x+5=0\). This simplifies to: \(-3x^{2}+3x+5=0\)

Step 4: Factor the equation

This is now a quadratic equation that can be factored. The factored equation is: \(-3x^{2}+3x+5=(-3x-5)(x-1)\)

Step 5: Solve for x

We now solve: -3x - 5 = 0 and x - 1 = 0. We find x = \(\frac{5}{3}\) and x = 1. However, if we substitute x = 1 to our original expression, the denominator becomes zero, which is undefined. Therefore, x = 1 is an extraneous solution and should be discarded

Step 6: Check the solution

We substitute x = \(\frac{5}{3}\) into the original equation to verify it is correct: \(5/( (\frac{5}{3})^2 - \frac{5}{3}) + 3/( \frac{5}{3} -1)=6\), and it simplifies to 6=6, which confirms our solution is correct

Key Concepts

Quadratic Equations

In mathematics, quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). These equations are pivotal in algebra and have numerous applications in science and engineering.
You might wonder how to recognize a quadratic equation in various contexts, especially when dealing with rational equations. Often, through some algebraic manipulation, such as eliminating denominators or combining terms, an equation transforms into a quadratic form.

• Quadratic equations are characterized by the term \(x^2\).
• They generally have two solutions, but these solutions could be real or complex numbers.
• The solutions to quadratic equations can be found using methods like factoring, completing the square, or the quadratic formula.

When solving rational equations, reaching a quadratic equation suggests that further algebraic techniques are needed, setting the stage for techniques like factoring to find the solutions.

Factoring

Factoring is a powerful algebraic technique used to simplify expressions and solve equations. When you factor a quadratic equation like \(-3x^2 + 3x + 5 = 0\), you essentially express it as a product of its linear factors.
The goal of factoring is to break down complex polynomial equations into simpler multipliers. This process makes it easier to find the roots of an equation.

• Look for common factors in each term first. Simplify the equation by dividing each term by these common factors.
• Recognize patterns such as the difference of squares or trinomials that allow for straightforward factoring.
• In challenging cases, use trial and error or algebraic identities to factor the equation.

For the problem given, the equation was factored into \((-3x-5)(x-1)\). This allowed for identifying potential solutions by setting each factor equal to zero.

Extraneous Solutions

When solving equations, especially complex ones involving fractions or rational expressions, you might encounter solutions that don't fit when substituted back into the original equation. These are known as extraneous solutions.
These solutions often arise when transformations, like multiplying by a common denominator, introduce possible roots that satisfy the manipulated equation but not the original one.

• Always check each potential solution by substituting it back into the original equation to identify extraneous solutions.
• Extraneous solutions are critical to consider because they don't satisfy the initial conditions of the problem.
• In the exercise provided, the solution \(x = 1\) turned out to be extraneous because it created an undefined expression when substituted back.

Being vigilant about extraneous solutions helps ensure the accuracy of your final answers and prevents misunderstandings in algebraic problems.