Problem 35 - Intermediate Algebra

Question

Solve each proportion. $$ \frac{2 b}{b+5}=\frac{-b}{3 b+8} $$

Step-by-Step Solution

Answer

The solutions are \(b = 0\) and \(b = -3\).

Step 1: Cross Multiply

To solve the proportion \( \frac{2b}{b+5} = \frac{-b}{3b+8} \), start by cross-multiplying to eliminate the fractions. Multiply the numerator of one fraction by the denominator of the other: \(2b \times (3b + 8) = -b \times (b + 5)\).

Step 2: Distribute Terms

Distribute in each multiplication. For the left side: \( 2b \times 3b = 6b^2 \) and \(2b \times 8 = 16b \), giving \( 6b^2 + 16b \). For the right side: \(-b \times b = -b^2\) and \(-b \times 5 = -5b\), giving \(-b^2 - 5b\).

Step 3: Combine All Terms

Set the equation from both sides together: \( 6b^2 + 16b = -b^2 - 5b \). Now, bring all terms to one side to set up the equation to zero: \(6b^2 + 16b + b^2 + 5b = 0\). Simplify to form: \(7b^2 + 21b = 0\).

Step 4: Factor Equation

Factor out the greatest common factor from each term: \(b(7b + 21) = 0\). This results in two possible solutions; setting each factor equal to zero: \(b = 0\) or \(7b + 21 = 0\).

Step 5: Solve for b

First, \(b = 0\) is one solution. For the second factor: \(7b + 21 = 0\), solve for \(b\): Subtract 21 from both sides to get \(7b = -21\), then divide by 7, resulting in \(b = -3\).

Key Concepts

Cross Multiplication

Cross multiplication is a helpful technique used to solve proportions. A proportion is an equation that states two ratios are equal, like \( \frac{2b}{b+5} = \frac{-b}{3b+8} \). By cross multiplying, you eliminate the fractions, making equations easier to solve. In this method, you multiply the numerator of one fraction by the denominator of the other. This means for our example: multiply \(2b\) by \((3b + 8)\) and \(-b\) by \((b + 5)\). It simplifies your problem from involving fractions to one using only simple multiplications and results in \(2b \cdot (3b + 8) = -b \cdot (b + 5)\). This step is pivotal because it sets the stage for solving the equation without dealing fractions directly.

Factoring Quadratics

Factoring quadratics can seem tricky, but it becomes manageable with practice. It's a critical skill in algebra because it helps simplify equations, making them solvable. For equations like \(7b^2 + 21b = 0\), you look for terms common in every part of the equation. Factoring involves identifying and separating these common factors. Here, both terms share a \(b\), so you can factor it out: \(b(7b + 21) = 0\).
This reduction opens paths to find solutions by setting the factors to zero, leading to equations like \(b = 0\) or \(7b + 21 = 0\). Recognizing the greatest common factor is significant because it transforms a complex quadratic into simpler linear equations.

Distributive Property

The distributive property is a fundamental concept that helps in simplifying algebraic expressions. It states that \( a(b + c) = ab + ac \). This property is used to ensure each term inside parentheses gets multiplied by the term outside. In our problem, after applying cross multiplication, you had expressions like \(2b(3b + 8)\). By distributing, you transform this into \(6b^2 + 16b\). Similarly, \(-b(b + 5)\) becomes \(-b^2 - 5b\).
This step is crucial because it breaks down complex expressions into simpler terms, paving the way to keep moving forward in solving the equation. Distributing correctly ensures no part of your expression is left out of calculations, maintaining balance in your algebraic process.

Algebraic Equation Solving

Solving algebraic equations involves finding values that satisfy the equation. Once you have a simplified equation like \(7b^2 + 21b = 0\), you’ll need to solve for \(b\). Start by factoring, bringing each expression into a simple format you recognize, and then set each factor equal to zero. For instance, from \(b(7b + 21) = 0\), deduce possible solutions by creating simple equations: \(b = 0\) or \(7b + 21 = 0\).
Then solve each smaller equation to find \(b\). For \(7b + 21 = 0\), subtract 21 and divide by 7, getting \(b = -3\). These factors each provide valid solutions for \(b\). This consistent approach simplifies the problem-solving process, allowing you to focus on manipulating and interpreting your equation correctly.