Problem 105 - Intermediate Algebra

Question

Solve: \(\left(\frac{1}{2}\right)^{-1}=\frac{5 b^{-1}}{2}+2 b(b+1)^{-1}\)

Step-by-Step Solution

Answer

The solution to the equation is \( b = -5 \).

Step 1: Simplify the Left Side of the Equation

The left side of the equation is \( \left(\frac{1}{2}\right)^{-1} \). The negative exponent indicates a reciprocal, so \( \left(\frac{1}{2}\right)^{-1} = 2^1 = 2 \). Therefore, the equation becomes \( 2 = \frac{5 b^{-1}}{2} + 2b(b+1)^{-1} \).

Step 2: Simplify the Terms on the Right Side

Simplify each term on the right side of the equation separately:1. \( \frac{5 b^{-1}}{2} \) becomes \( \frac{5}{2b} \) because \( b^{-1} = \frac{1}{b} \).2. \( 2b(b+1)^{-1} \) becomes \( \frac{2b}{b+1} \) as \( (b+1)^{-1} = \frac{1}{b+1} \).Thus, the equation is now \( 2 = \frac{5}{2b} + \frac{2b}{b+1} \).

Step 3: Eliminate the Fractions

To eliminate the fractions, find a common denominator. The denominators are \( 2b \) and \( b+1 \) which combines using the least common multiple to be \( 2b(b+1) \).Multiply the entire equation by \( 2b(b+1) \) to clear the fractions:\[ 2 \cdot 2b(b+1) = \left(\frac{5}{2b}\right) \cdot 2b(b+1) + \left(\frac{2b}{b+1}\right) \cdot 2b(b+1) \]Simplifying each term results in:\[ 4b(b+1) = 5(b+1) + 2b(2b) \]

Step 4: Simplify and Solve for \( b \)

Distribute and simplify each term:- Left side: \( 4b^2 + 4b \)- Right side: \( 5b + 5 + 4b^2 \)Setting them equal gives:\[ 4b^2 + 4b = 4b^2 + 5b + 5 \]Subtracting \( 4b^2 \) from both sides results in:\[ 4b = 5b + 5 \]Now, subtract \( 5b \) from both sides:\[ 4b - 5b = 5 \]\[ -b = 5 \]Thus, \( b = -5 \).

Key Concepts

Understanding Negative Exponents

Negative exponents might look confusing at first, but they're really just another way to express division or reciprocals. When you see a negative exponent, it means you need to take the reciprocal of the base and raise it to the positive of that exponent. For example,

• The expression \( a^{-1} \) is equivalent to \( \frac{1}{a} \).
• Therefore, \( \left(\frac{1}{2}\right)^{-1} \) becomes \( 2^1 \), which is simply 2.

Breaking down negative exponents this way always works. Just flip the base (find the reciprocal) and make the exponent positive. By handling negative exponents correctly, equations often become much simpler to solve.

Reciprocals in Equations

Reciprocals play a vital role in manipulating and simplifying expressions, especially when negative exponents are involved. Understanding the reciprocal of a number involves flipping the number:

• The reciprocal of a fraction \( \frac{x}{y} \) is \( \frac{y}{x} \).
• For a whole number \( n \), its reciprocal is \( \frac{1}{n} \).

For example, in the expression \( b^{-1} \), it's reciprocated to give \( \frac{1}{b} \). Recognizing reciprocals in equations assists in making terms straightforward, setting a solid foundation for further simplification. Being adept with reciprocals accelerates simplifying any equation.

Techniques for Clearing Fractions

Clearing fractions in an equation is a key step to simplify the equation, making it easier to work with. To clear fractions, follow these steps:

• Identify all the fractions in the equation and determine their denominators.
• Find the least common multiple (LCM) of these denominators to establish a common ground to eliminate fractions.
• Multiply every term of the equation by this LCM. This removes the fractions in the equation by canceling out the denominators.

In our exercise, the denominators \( 2b \) and \( b+1 \) are combined to form \( 2b(b+1) \). Multiplying each term by this clears the fractions efficiently, which simplifies further processing of the equation.

Solving for Variables After Simplification

Once fractions are cleared, and the equation is in a more manageable form, solving for the variable becomes the main goal. This usually involves:

• Distributing and combining like terms on both sides of the equation.
• Isolating the variable by moving all terms involving it to one side, and constants to the other.
• Simplifying both sides, ensuring the variable is explicitly solved.

For instance, with the equation \( 4b^2 + 4b = 4b^2 + 5b + 5 \), subtract any duplicate terms like \( 4b^2 \) to trim the equation down. Once simplified, solve for \( b \) by isolating it to give \( b = -5 \). Each step brings you closer to the solution, reinforcing problem-solving skills every time you simplify an expression and solve for variables.