Problem 53 - Basic Technical Mathematics with Calculus

Question

Solve the given equations involving fractions. $$\frac{1}{x-3}+\frac{4}{x}=2$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{3}{2}\) and \(x = 4\).

Step 1: Identify the Least Common Denominator (LCD)

The denominators in the equation are \(x-3\) and \(x\). The least common denominator is the product of these two expressions, which is \((x-3)x\). We will use the LCD to combine the fractions.

Step 2: Clear the Fractions

Multiply every term by the LCD \((x-3)x\) to eliminate the fractions: \[(x-3)x \left( \frac{1}{x-3} \right) + (x-3)x \left( \frac{4}{x} \right) = 2(x-3)x\] This simplifies to: \[x + 4(x-3) = 2x(x-3)\]

Step 3: Simplify and Expand

Distribute and simplify the equation:\[x + 4(x-3) = 2x^2 - 6x\]Expanding \(4(x-3)\) we get \[4x - 12\]. Thus:\[x + 4x - 12 = 2x^2 - 6x\]Combine like terms to get:\[5x - 12 = 2x^2 - 6x\]

Step 4: Move All Terms to One Side

Rearrange the equation to set it to zero:\[0 = 2x^2 - 6x - 5x + 12\]This simplifies to:\[0 = 2x^2 - 11x + 12\]

Step 5: Factor the Quadratic Equation

Factor the quadratic equation \(2x^2 - 11x + 12\). We look for factors of \(24\) (product of \(2\) and \(12\)) that sum to \(-11\). These factors are \(-3\) and \(-8\). Rewriting the middle term:\[2x^2 - 3x - 8x + 12 = 0\]Group terms:\[(2x^2 - 3x) - (8x - 12) = 0\]Factor each group:\[x(2x - 3) - 4(2x - 3) = 0\]Factor out the common factor \((2x - 3)\):\[(2x - 3)(x - 4) = 0\]

Step 6: Solve for x

Set each factor equal to zero:\[2x - 3 = 0\] or \[x - 4 = 0\]Solving these gives:\[2x = 3 \implies x = \frac{3}{2}\]\[x = 4\]

Step 7: Check for Extraneous Solutions

Verify that the solutions do not make any original denominators equal zero. Substituting into \(x-3\) and \(x\) verifies:For \(x = \frac{3}{2}\), both \((\frac{3}{2}-3)\) and \(\frac{3}{2}\) are non-zero.For \(x = 4\), both \((4-3)\) and \(4\) are non-zero.Thus, both solutions are valid.

Key Concepts

Least Common Denominator

When working with fraction equations, like \( \frac{1}{x-3}+\frac{4}{x}=2 \), knowing how to find the Least Common Denominator (LCD) is essential. The LCD is the smallest expression that all denominator terms can divide into without a remainder. In our equation, the denominators are \( x-3 \) and \( x \). These expressions do not share common factors, so their product \((x-3)x\) becomes the LCD.

Why is finding the LCD important? Using the LCD allows us to clear the fractions by transforming the entire equation. This step simplifies solving the equation because fractions can be tricky to work with. By multiplying each term by the LCD, we eliminate denominators. This transforms our original fraction-heavy equation into a friendlier, fraction-free equation.

Quadratic Equation

After clearing the fractions in our original problem, we obtain a new equation: \(5x - 12 = 2x^2 - 6x \). This equation is a quadratic equation, recognized by its highest term, \(x^2\). Quadratic equations are common in algebra and have the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants.

Quadratic equations can be solved in multiple ways, including factoring, completing the square, or using the quadratic formula. This particular problem involves moving all terms to one side, giving us \(0 = 2x^2 - 11x + 12 \). This arrangement sets up an equation perfectly for factoring, a process that breaks down the equation for simpler solution finding.

Factoring Method

Factoring is a vital skill in solving quadratic equations like \(2x^2 - 11x + 12 = 0 \). The goal is to express the quadratic expression as a product of two binomials. Here, we factor by first identifying two numbers that multiply to the product of the leading coefficient and constant term (\(2 \times 12 = 24\)) and add up to the middle coefficient (\(-11\)).

Once we find these numbers, \(-3\) and \(-8\), we can rewrite the middle term, resulting in \(2x^2 - 3x - 8x + 12\). By grouping terms, \((2x^2 - 3x) - (8x - 12)\), and factoring each group, the quadratic beautifully factors to \((2x - 3)(x - 4) = 0\).

This factorization allows us to set each factor equal to zero in order to solve for \(x\). Such techniques are not only applicable in algebraic problems, but essential as you progress into more advanced mathematics.