Problem 28 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The solutions are \(x = 3\) and \(x = -\frac{2}{5}\).

Step 1: Determine Common Denominator

To solve the equation \( \frac{4}{x+1} + \frac{2}{x} = \frac{5}{3} \), we first need a common denominator for the fractions on the left side. The denominators are \(x+1\) and \(x\), so the common denominator would be \((x+1)x\).

Step 2: Rewrite Each Fraction

Express each term with the common denominator: \[ \frac{4}{x+1} = \frac{4x}{x(x+1)} \]\[ \frac{2}{x} = \frac{2(x+1)}{x(x+1)} \].Now the equation becomes: \[ \frac{4x + 2(x+1)}{x(x+1)} = \frac{5}{3} \].

Step 3: Simplify the Numerator

Expand and simplify the numerator: \(4x + 2(x+1)\) becomes \(4x + 2x + 2 = 6x + 2\). The equation now looks like: \[ \frac{6x + 2}{x(x+1)} = \frac{5}{3} \].

Step 4: Cross-Multiply

Cross-multiply to eliminate the fractions. This results in: \[ 3(6x + 2) = 5x(x+1) \].

Step 5: Expand and Simplify Both Sides

Expand both sides of the equation:Left side: \(3 \times 6x + 3 \times 2 = 18x + 6\).Right side: \(5x^2 + 5x\).Resulting in the equation: \[ 18x + 6 = 5x^2 + 5x \].

Step 6: Rearrange Into a Standard Quadratic Form

Move all terms to one side to form a quadratic equation:\[ 5x^2 + 5x - 18x - 6 = 0 \].Simplify to get:\( 5x^2 - 13x - 6 = 0 \).

Step 7: Use the Quadratic Formula

Apply the quadratic formula:\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \] where \(a = 5\), \(b = -13\), and \(c = -6\).Calculate the discriminant: \[ (-13)^2 - 4 \times 5 \times (-6) = 169 + 120 = 289 \].The formula gives:\[ x = \frac{{13 \pm \sqrt{289}}}{10} \].Since \(\sqrt{289} = 17\), the solutions are: \[ x = \frac{{13 + 17}}{10} \] and \[ x = \frac{{13 - 17}}{10} \].

Step 8: Calculate the Roots

For \(x = \frac{13 + 17}{10} = 3\).For \(x = \frac{13 - 17}{10} = -\frac{2}{5}\).

Key Concepts

Common Denominator

When working with fractions, especially in equations, finding a common denominator is essential. This process simplifies the addition or subtraction of fractions, making calculations straightforward. In the provided equation, \( \frac{4}{x+1} + \frac{2}{x} = \frac{5}{3} \), the denominators \( x+1 \) and \( x \) have been identified for the fractions on the left-hand side. The concept of a common denominator is to transform these separate fractions into expressions with the same denominator, allowing for easy combination. To find a common denominator, simply multiply the different denominators together. That gives us the least common multiple for our case, \( (x+1)x \). The expression can now be rewritten as:

• \( \frac{4}{x+1} \rightarrow \frac{4x}{x(x+1)} \)
• \( \frac{2}{x} \rightarrow \frac{2(x+1)}{x(x+1)} \)

Both fractions now share a common denominator \( x(x+1) \). This step crucially sets up the required stage for simplifying, solving, or further manipulation of the equation.

Cross-Multiplication

Cross-multiplication is a useful technique when dealing with fractions, particularly in equations involving them. It eliminates the fractions by multiplying the numerator of one fraction by the denominator of the other, ensuring the equation converts into a more familiar algebraic form. In our example, the equation \( \frac{6x + 2}{x(x+1)} = \frac{5}{3} \) requires us to cross-multiply to solve for \( x \). Here’s how it works:

• Multiply the numerator on the left by the denominator on the right: \( 3(6x + 2) \)
• Multiply the numerator on the right by the denominator on the left: \( 5x(x+1) \)

This gives us the equation \( 3(6x + 2) = 5x(x+1) \).Cross-multiplication helps us switch from fractions to a more traditional algebraic expression, making it possible to manage and simplify complicated equations easily.

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations, taking the general form of \( ax^2 + bx + c = 0 \). It provides a straightforward method to find the roots of nearly any quadratic equation.For the equation \( 5x^2 - 13x - 6 = 0 \) that results from our example after simplifying and rearranging, apply the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Where the coefficients are

• \( a = 5 \)
• \( b = -13 \)
• \( c = -6 \)

Calculate the discriminant: \(-13^2 - 4 \times 5 \times (-6) = 169 + 120 = 289\).Since the discriminant (a critical component determining the nature of the roots) is positive, we find real and distinct solutions. Solving gives us:

• \( x = \frac{13 + 17}{10} = 3 \)
• \( x = \frac{13 - 17}{10} = -\frac{2}{5} \)

The quadratic formula elegantly provides a means to calculate the roots of any quadratic equation swiftly and correctly.