Problem 54 - Basic Technical Mathematics with Calculus

Question

Solve the given equations involving fractions. $$2-\frac{1}{x}=\frac{3}{x+2}$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 1\) and \(x = -1\).

Step 1: Isolate the Fraction with x

First, we need to isolate the term with the fraction \( \frac{1}{x} \). Subtract 2 from both sides of the equation:\[-\frac{1}{x} = \frac{3}{x+2} - 2\]

Step 2: Express as a Single Fraction

We will convert \( 2 \) into a fraction with denominator \( x+2 \) so that we can perform the subtraction easily:\[-\frac{1}{x} = \frac{3}{x+2} - \frac{2(x+2)}{x+2}\]

Step 3: Simplify the Right Side

Distribute and subtract on the right side:\[\frac{3}{x+2} - \frac{2(x+2)}{x+2} = \frac{3 - 2x - 4}{x+2} = \frac{-2x - 1}{x+2}\]Now, the equation becomes:\[-\frac{1}{x} = \frac{-2x - 1}{x+2}\]

Step 4: Eliminate Fractions by Cross-Multiplying

Eliminate the fractions by cross-multiplying:\[(x)(-2x - 1) = (-1)(x+2)\]Which simplifies to:\[-2x^2 - x = -x - 2\]

Step 5: Combine Like Terms

Move all terms to one side of the equation:\[-2x^2 - x + x + 2 = 0\]Simplify to:\[-2x^2 + 2 = 0\]

Step 6: Solve the Quadratic Equation

To solve for \(x\), first add \(2x^2\) to both sides:\[2x^2 = 2\]Divide both sides by 2:\[x^2 = 1\]Take the square root of both sides to find both solutions:\[x = \, \pm 1\]

Step 7: Check Solutions for Validity

We need to ensure that these solutions do not cause division by zero in the original equation. Substitute back to check:- For \(x = 1\): The original denominators \(x = 1\) and \(x + 2 = 3\) are valid (non-zero).- For \(x = -1\): Both denominators \(x = -1\) and \(x + 2 = 1\) are valid.Thus, both \(x = 1\) and \(x = -1\) are solutions.

Key Concepts

Cross-Multiplying Fractions

Cross-multiplying is a useful technique to eliminate fractions from an equation, making it easier to solve. It involves multiplying the numerator of one fraction by the denominator of the other and vice versa, then setting the two products equal to each other. This technique transforms two fractions into a single equation without fractions to handle. Here’s a quick breakdown:

• Given two fractions: \( \frac{a}{b} = \frac{c}{d} \)
• Cross-multiply to get: \( a \cdot d = b \cdot c \)

Once you've cross-multiplied, you can solve the resulting equation using standard algebraic methods.
In the exercise, cross-multiplication helped eliminate fractions by equating the product of the numerator and denominator with each other, turning the problem into a quadratic equation for easier solving.

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \). They can be solved using several methods, including factoring, completing the square, or using the quadratic formula. Each method has different applications depending on the form and complexity of the quadratic equation.To solve the quadratic equation attained in the exercise, \(-2x^2 + 2 = 0\), follow these steps:

• First, isolate the squared term: \( 2x^2 = 2 \).
• Then, divide all terms by the coefficient of \( x^2 \): \( x^2 = 1 \).
• Finally, find the square root of both sides to solve for \( x \), giving \( x = \pm 1 \).

Quadratic equations like this can yield up to two real solutions, as seen in \( x = 1 \) and \( x = -1 \). Don't forget to verify these solutions in the context of the original problem to ensure they do not result in any mathematical errors, such as division by zero.

Simplifying Algebraic Expressions

Simplifying algebraic expressions involves combining like terms and reducing fractions to their simplest form, making equations easier to work with. This process can include distributing multiplication over addition, combining constants, and reducing fractions by finding common factors.The exercise demonstrated simplification through subtraction of fractions:

• Convert constants to fractions with common denominators, such as \( 2 \) to \( \frac{2(x+2)}{x+2} \).
• Perform the arithmetic on the numerators while keeping the denominator constant.
• Simplify the resulting expression by performing arithmetic operations and combining like terms: \( \frac{3 - 2x - 4}{x+2} = \frac{-2x - 1}{x+2} \).

Effective simplification clears a path to more straightforward computation, leading to easily solvable equations, as shown in further steps. Understanding this process helps manage complicated algebraic expressions and solve equations efficiently.