Problem 60 - Intermediate Algebra

Question

Solve each proportion. $$ \frac{b^{2}}{5}=\frac{b}{6 b-13} $$

Step-by-Step Solution

Answer

The solutions are \( b = \frac{5}{2} \) and \( b = -\frac{1}{3} \).

Step 1: Identify the Proportion

We have the proportion \( \frac{b^2}{5} = \frac{b}{6b-13} \). We need to find the value of \( b \) without any undefined operations.

Step 2: Check for Undefined Values

The fraction \( \frac{b}{6b-13} \) becomes undefined if the denominator, \( 6b-13 \), equals zero. Therefore, solve for \( b \) in \( 6b-13 = 0 \) to find \( b = \frac{13}{6} \). Since \( b = \frac{13}{6} \) would make the denominator zero, it must be excluded from the solution set.

Step 3: Cross-Multiply to Eliminate Fractions

We can eliminate the fractions by cross-multiplying: \( b^2 \cdot (6b - 13) = 5b \). This results in the equation \( 6b^3 - 13b^2 = 5b \).

Step 4: Simplify the Equation

Bring all terms to one side of the equation: \( 6b^3 - 13b^2 - 5b = 0 \). Factor out the common term \( b \): \( b(6b^2 - 13b - 5) = 0 \).

Step 5: Solve for b

We have \( b = 0 \) or \( 6b^2 - 13b - 5 = 0 \). For \( 6b^2 - 13b - 5 = 0 \), use the quadratic formula: \( b = \frac{-(-13) \pm \sqrt{(-13)^2 - 4 \times 6 \times (-5)}}{2 \times 6} \).

Step 6: Apply the Quadratic Formula

Calculate the discriminant: \( 169 + 120 = 289 \), and \( \sqrt{289} = 17 \). So, \( b = \frac{13 \pm 17}{12} \). This leads to solutions \( b = \frac{30}{12} = \frac{5}{2} \) and \( b = \frac{-4}{12} = -\frac{1}{3} \).

Step 7: Validate the Solutions

The fraction \( \frac{b}{6b-13} \) remains valid as long as the denominator is not zero. For \( b = \frac{5}{2} \), \( 6b - 13 e 0 \); for \( b = -\frac{1}{3} \), \( 6b - 13 e 0 \). Hence, both \( b = \frac{5}{2} \) and \( b = -\frac{1}{3} \) are valid solutions.

Key Concepts

Cross-multiplication

Cross-multiplication is a trusty method used to solve equations that have proportions, or fractions, on either side. When faced with, let's say, two fractions set equal to one another, we can "cross-multiply" to clear those pesky fractions right off the table.

This works by taking the diagonal terms and multiplying them across the equation. So, if you start with a proportion like \( \frac{a}{b} = \frac{c}{d} \), you can transform it into \( a \cdot d = b \cdot c \).

• Multiply the numerator of the first fraction by the denominator of the second.
• Multiply the numerator of the second fraction by the denominator of the first.

In the original exercise, the equation \( \frac{b^2}{5} = \frac{b}{6b-13} \) was converted into \( b^2 \cdot (6b - 13) = 5b \) through cross-multiplication. This makes further steps simpler by creating a single equation to solve for \( b \). This method is powerful because it allows you to focus on a straightforward equation without the complications that fractions can introduce.

Undefined values

In any mathematical expression, fractions become tricky when their denominators hit the forbidden number zero. When the denominator of a fraction is zero, the value becomes undefined; this is a critical point and must always be checked first.

To find these undefined values, look at the denominator and set it to zero, just like in our case: \( 6b-13 \).

• Solve \( 6b - 13 = 0 \) to identify potential problematic values.
• Calculate to isolate \( b \), giving \( b = \frac{13}{6} \).

Since \( b = \frac{13}{6} \) makes the denominator zero, this value is not valid in the equation and must be excluded from the solution set. Understanding undefined values is crucial because ignoring them can lead to incorrect solutions that cause logical errors in mathematical reasoning.

Quadratic formula

The quadratic formula is like a loyal magician in algebra, always ready to find solutions for any quadratic equation. A quadratic equation is typically in the form \( ax^2 + bx + c = 0 \) and holds up to two possible solutions.

• The formula is: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Calculate the discriminant \( b^2 - 4ac \), which suggests the nature of the solutions.
• Substitute the values for \( a \), \( b \), and \( c \) into the formula.

For our exercise, you rearranged \( 6b^2 - 13b - 5 = 0 \) and used the quadratic formula to find the values of \( b \).

By substituting \( a = 6 \), \( b = -13 \), and \( c = -5 \), the resulting values were \( b = \frac{5}{2} \) and \( b = -\frac{1}{3} \). The quadratic formula is incredibly handy when factoring sounds like a daunting task, particularly for complex numbers or odd coefficients.