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Chapter 6: Problem 8

Find each product or quotient. $$\frac{1}{\square} \cdot \frac{x}{1(x+1)}=\frac{1}{x}$$

Short Answer

Expert verified
The equation does not have a solution, as it contradicts itself, resulting in 0 = 1 in the process of solving for x.

Step by step solution

01

Simplify the expression

First, replace the square with x. The equation becomes \(\frac{1}{x} \cdot \frac{x}{x+1}=\frac{1}{x}\). If we multiply \(\frac{1}{x}\) with \(\frac{x}{x+1}\), the x's cancel out in the numerator and denominator, leading to \(\frac{1}{x+1}=\frac{1}{x}\).
02

Cross multiply

Cross multiply to remove the fraction and get the equation in simple form. For \(\frac{1}{x+1}=\frac{1}{x}\), cross-multiplication gives \(x = x+1\).
03

Solve for x

To find the value for x, isolate x. In \(x = x+1\), moving \(x\) over to the right side gives us the equation 0 = 1, which is not possible.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross Multiplication
Cross multiplication is a handy technique used to solve equations that involve two fractions. It essentially involves multiplying the numerator of one fraction by the denominator of the other. This helps eliminate the fractions and turns the equation into a simpler algebraic expression, making it much easier to solve.
To apply cross multiplication effectively, here's what you鈥檇 do:
  • Identify the equation where two fractions are set equal to each other, for example, \( \frac{a}{b} = \frac{c}{d} \).
  • Cross multiply by multiplying across diagonals. This means you multiply \( a \cdot d \) and \( b \cdot c \), resulting in the equation \( a \cdot d = b \cdot c \).
This step simplifies the equation into a format where fractions are no longer present, allowing for straightforward algebraic solving. Remember, cross multiplication is only valid when you have a fraction on both sides of the equation.
Solving Equations
Solving equations is all about finding the value of the unknown variable that makes the equation true. Once you've simplified the equation using techniques like cross multiplication, you'll usually end up with a straightforward algebraic equation.
Here's how you proceed to solve for the variable:
  • Bring all terms involving the variable to one side of the equation and keep constant values on the opposite side.
  • Simplify both sides of the equation as needed. This could involve combining like terms or simply arithmetic operations.
  • Isolate the variable by using operations such as addition, subtraction, multiplication, or division.
In the exercise given, once you reach the equation \(x = x+1\), you'll find that it's unsolvable in the real number system because it implies \(0 = 1\). This indicates no solution exists for the equation, or there might have been a mistake or assumption in setting up the equation.
Simplifying Fractions
Simplifying fractions is a crucial step in handling complex algebraic expressions. It involves reducing the fraction to its lowest possible form. This is achieved by dividing both the numerator and the denominator by their greatest common divisor.
For algebraic fractions, like the ones in the problem, follow these steps:
  • Check for common factors in each part of the fraction, particularly terms or variables that appear in both the numerator and the denominator.
  • Cancel these common factors, simplifying the fraction to its irreducible form.
For instance, if you have \( \frac{x}{x+1} \), you look for common factors. In this case, cancelling across the terms isn鈥檛 possible unless \(x = x+1\), which results directly in contradiction as seen in our solved step-by-step exercise. Simplifying appropriately ensures that further algebraic manipulations are easier and error-free.

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