Problem 28 - Algebra for College Students

Question

Solve each equation. $$\frac{4}{x+1}+\frac{2}{x}=\frac{5}{3}$$

Step-by-Step Solution

Answer

The solution is \( x = 3 \).

Step 1: Find a Common Denominator

The common denominator for the fractions \( \frac{4}{x+1} \), \( \frac{2}{x} \), and \( \frac{5}{3} \) is \( (x+1)x3 \). We will multiply each term by this common denominator to eliminate the fractions.

Step 2: Multiply Through by the Common Denominator

Multiply each term of the equation by the common denominator:\[(x+1)x3 \left( \frac{4}{x+1} + \frac{2}{x} = \frac{5}{3} \right)\]This results in:\[4x3 + 2(x+1)3 = 5(x)(x+1)\]

Step 3: Simplify Each Term

Simplify each term by canceling the denominators:\[12x + 6(x+1) = 5x(x+1)\]

Step 4: Distribute and Combine Like Terms

Distribute and simplify:\[12x + 6x + 6 = 5x^2 + 5x\]Combine like terms:\[18x + 6 = 5x^2 + 5x\]

Step 5: Rearrange into a Quadratic Equation

Move all terms to one side to form a quadratic equation:\[0 = 5x^2 - 13x - 6\]This simplifies to:\[5x^2 - 13x - 6 = 0\]

Step 6: Solve the Quadratic Equation

Solve the quadratic equation using the quadratic formula, where \( a = 5 \), \( b = -13 \), and \( c = -6 \).\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Calculate the discriminant:\[ b^2 - 4ac = (-13)^2 - 4(5)(-6) = 169 + 120 = 289 \]Now substitute into the formula:\[ x = \frac{13 \pm \sqrt{289}}{10} \]\[ x = \frac{13 \pm 17}{10} \]

Step 7: Calculate Solutions

The roots are:1. \( x = \frac{13 + 17}{10} = 3 \)2. \( x = \frac{13 - 17}{10} = -\frac{2}{5} \)

Step 8: Verify Solutions within Original Equation

Check which solutions satisfy the original equation:Re-substitute \( x = 3 \):\[ \frac{4}{3+1} + \frac{2}{3} = \frac{5}{3} \]This is true. Now check \( x = -\frac{2}{5} \):\[ \frac{4}{-rac{2}{5}+1} + \frac{2}{-rac{2}{5}} \]Leads to division by zero, hence invalid.Thus, only \( x = 3 \) is valid.

Key Concepts

Common Denominator

When dealing with fractions in an equation, a common denominator helps in simplifying the process. The common denominator is a shared multiple of the denominators in each fraction. In our example, we have the fractions \( \frac{4}{x+1} \), \( \frac{2}{x} \), and \( \frac{5}{3} \). The common denominator is determined by multiplying together each unique term in the denominators, yielding \((x+1)x3\).

By multiplying each term of the equation by this common denominator, we eliminate the fractions. This makes the equation easier to manipulate and solve. The ability to identify and use a common denominator is a crucial skill for solving equations involving fractions.

Quadratic Equation

A quadratic equation is a polynomial equation of degree two, generally in the form \( ax^2 + bx + c = 0 \). In mathematical terms, it illustrates a parabola when graphed.
In the solution provided, after simplifying and rearranging the equation, we have: \( 5x^2 - 13x - 6 = 0 \). This format makes it easy to apply algebraic methods to find solutions.

Understanding how to transform an equation into a standard quadratic form makes solving complex equations more approachable. Often, these equations arise from real-world problems like projectile motion, making their comprehensive knowledge essential in various fields.

Quadratic Formula

The quadratic formula provides a solution for quadratic equations and is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula allows us to find the roots of any quadratic equation provided the discriminant \( b^2 - 4ac \) is non-negative. In our case, the discriminant is 289, which is perfectly positive, permitting real solutions. We apply the formula using values \( a = 5 \), \( b = -13 \), and \( c = -6 \), finally yielding the solutions \( x = 3 \) and \( x = -\frac{2}{5} \).

Mastering the quadratic formula empowers you with a tool to tackle a significant category of algebraic problems.
It is one of the fundamental methods taught for solving quadratic equations.

Verify Solutions

Verifying solutions is a crucial step that ensures the solutions satisfy the original equation. In the exercise, after calculating the roots \( x = 3 \) and \( x = -\frac{2}{5} \), we substitute these back into the original equation to confirm their validity.

Upon testing, we find:

• \( x = 3 \) holds true within the equation, ensuring it's a valid solution.
• \( x = -\frac{2}{5} \) results in division by zero, rendering it invalid.

Verifying solutions often reveals any extraneous roots and confirms the accuracy of your solutions. This step prevents bringing inaccurate roots to further calculations or practical applications, thereby cementing the solved equation's accuracy and reliability.