Problem 24 - Elementary Algebra

Question

$$ \text { For Problems 1-32, solve each equation. (Objective 1) } $$ $$ \frac{10}{7 x}+\frac{10}{7(x+3)}=1 $$

Step-by-Step Solution

Answer

The solution is \( x = 2 \).

Step 1: Identify the Least Common Denominator

To solve the equation \( \frac{10}{7x} + \frac{10}{7(x+3)} = 1 \), first find the least common denominator (LCD) for the fractions. The denominators are \( 7x \) and \( 7(x+3) \), so the LCD is \( 7x(x+3) \).

Step 2: Multiply Through by the LCD

Multiply every term in the equation by the LCD \( 7x(x+3) \) to eliminate the fractions:\[ 7x(x+3) \left( \frac{10}{7x} \right) + 7x(x+3) \left( \frac{10}{7(x+3)} \right) = 7x(x+3) \times 1 \]Simplify to get:\[ 10(x+3) + 10x = 7x(x+3) \]

Step 3: Expand and Simplify

Expand the left side:\[ 10x + 30 + 10x = 7x^2 + 21x \]Combine like terms:\[ 20x + 30 = 7x^2 + 21x \]

Step 4: Rearrange the Equation

Bring all terms to one side to form a quadratic equation:\[ 7x^2 + 21x - 20x - 30 = 0 \]Simplify to:\[ 7x^2 + x - 30 = 0 \]

Step 5: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 7 \), \( b = 1 \), and \( c = -30 \).Calculate the discriminant:\[ b^2 - 4ac = 1^2 - 4(7)(-30) = 1 + 840 = 841 \]The square root of 841 is 29.Calculate the roots:\[ x = \frac{-1 \pm 29}{14} \]

Step 6: Calculate Specific Roots

Calculate each root:First root: \( x = \frac{-1 + 29}{14} = 2 \).Second root: \( x = \frac{-1 - 29}{14} = -\frac{30}{14} = -\frac{15}{7} \).

Step 7: Verify Solutions

Check both solutions in the original equation:For \( x = 2 \):\[ \frac{10}{14} + \frac{10}{35} = 1 \quad \text{holds true} \].For \( x = -\frac{15}{7} \): the terms involve division by zero, so this solution is not valid.

Key Concepts

Least Common Denominator

When solving equations that involve fractions, one essential step is to clear the fractions by finding the Least Common Denominator (LCD). The LCD is the smallest expression that each of the denominators can divide into without a remainder. This allows you to eliminate the fractions and work with simpler terms.

In our problem, the denominators are \( 7x \) and \( 7(x+3) \). To find the LCD, multiply the distinct parts together: \( 7 \, \times \, x \, \times \, (x+3) = 7x(x+3) \). This expression serves as the common ground for both fractions.

Why is this step important? It simplifies the equation, making it easier to handle. By multiplying every term by the LCD, you clear the fractions, reducing a complex fractional equation to a more straightforward algebraic one.

Fractions

Fractions represent a part of a whole and are often present in algebraic equations. They consist of a numerator and a denominator, presenting ratios such as \( \frac{10}{7x} \). Understanding how to manipulate fractions is crucial.

When dealing with fractions in equations, consider the following approaches:

• Cross-Multiplication: Occasionally used for equations with a single fraction on each side.
• Clear Fractions: Multiply every term by the LCD to transform the equation into one without fractions.

In our exercise equation, fractions are effectively managed by multiplying with the LCD (\( 7x(x+3) \)) which helps in removing the denominators. This transforms the equation, allowing polynomial operations rather than fractional additions.

Quadratic Formula

The Quadratic Formula is a reliable method for finding the roots of a quadratic equation, which is an equation of the form \( ax^2 + bx + c = 0 \). This formula is especially useful when the equation does not factor easily.

The standard form of the Quadratic Formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where:

• \(a\) is the coefficient of \(x^2\)
• \(b\) is the coefficient of \(x\)
• \(c\) is the constant term

In our example, the equation \(7x^2 + x - 30 = 0\) requires solving for \(x\) using the quadratic formula because it is not easily factorable. Substitute \(a = 7\), \(b = 1\), and \(c = -30\) into the formula to calculate the roots.

An essential part of using the quadratic formula is calculating the discriminant \(b^2 - 4ac\), which indicates the nature of the roots. A perfect square gives two real rational roots, while a negative value would imply complex roots. Here, since the discriminant \(841\) is positive and a perfect square, it results in two real solutions: \(x = 2\) and \(x = -\frac{15}{7}\).