Problem 28 - Algebra A Combined Function

Question

Solve each equation. $$ \frac{5}{x-6}=\frac{x}{x-2} $$

Step-by-Step Solution

Answer

The solutions are \(x=1\) and \(x=10\).

Step 1: Identify the common denominators

The given equation is \( \frac{5}{x-6} = \frac{x}{x-2} \). The denominators are \(x-6\) and \(x-2\). The common denominator for the entire equation is \((x-6)(x-2)\).

Step 2: Cross-multiply to eliminate fractions

To solve the equation, multiply both sides by the common denominator \((x-6)(x-2)\) to eliminate the fractions. This gives:\[5(x-2) = x(x-6)\]

Step 3: Expand both sides

Distribute on both sides of the equation:\[5x - 10 = x^2 - 6x\]

Step 4: Rearrange the equation

Bring all terms to one side to set the quadratic equation to zero:\[0 = x^2 - 6x - 5x + 10\]Simplify this to:\[0 = x^2 - 11x + 10\]

Step 5: Factor the quadratic equation

Factor the quadratic equation \(x^2 - 11x + 10\). We are looking for two numbers that multiply to 10 and add up to -11. These numbers are -1 and -10. Thus:\[0 = (x-1)(x-10)\]

Step 6: Solve for x

Set each factor equal to zero:1. \(x-1=0\) gives \(x=1\)2. \(x-10=0\) gives \(x=10\)

Step 7: Check solutions

Substitute \(x=1\) and \(x=10\) back into the original equation to ensure they do not result in a zero denominator.- For \(x=1\): the denominator \(x-6 = -5\) and \(x-2 = -1\) are non-zero.- For \(x=10\): the denominator \(x-6 = 4\) and \(x-2 = 8\) are non-zero.Both values are valid solutions.

Key Concepts

Factoring Quadratics

When solving quadratic equations, factoring is a method used to express the quadratic in terms of simpler polynomials. This helps to find the roots or solutions of the equation. For the quadratic equation in our problem, we have:- Quadratic expression: \(x^2 - 11x + 10\)To factor this expression, we need to find two numbers that multiply to the constant term, which is 10, while simultaneously adding up to the linear coefficient, which is -11. In this case:- The product is 10.- The sum is -11.The numbers that satisfy these conditions are -1 and -10. Thus, the quadratic can be factored as:\[(x - 1)(x - 10)\]By setting each factor equal to zero, we find the potential solutions for \(x\), helping to determine where the original quadratic is equal to zero.

Cross-Multiplication in Equations

Cross-multiplication is a technique used to eliminate fractions in equations, typically when dealing with a proportion between two fractions. It involves multiplying the numerator of each fraction by the denominator of the other. In our exercise:- Given equation: \( \frac{5}{x-6} = \frac{x}{x-2} \)By cross-multiplying, we multiply 5 by \(x-2\) and \(x\) by \(x-6\). This helps simplify the equation and eliminate the fractions:- Multiply the numerator of the left side by the denominator of the right: \(5(x-2)\)- Multiply the numerator of the right side by the denominator of the left: \(x(x-6)\)This results in a simpler quadratic equation:\[5(x-2) = x(x-6)\]Through cross-multiplication, we can now solve the equation without dealing with fractions, making it easier to handle.

Identifying Common Denominators

Identifying common denominators is an essential skill when working with rational equations. It allows us to combine fractions by ensuring they have the same denominator, which simplifies solving them. In our given exercise:- Original equations: \(\frac{5}{x-6} = \frac{x}{x-2}\)The denominators are \(x-6\) and \(x-2\). To eliminate the fractions and work with a unified equation, find a common denominator. The common denominator is the product of the different denominators, which in this case is:- Common denominator: \((x-6)(x-2)\)With this common denominator, you can cross-multiply to remove fractions, leading to a linear or quadratic equation that can be solved using algebraic techniques like factoring. This process transforms a problematic rational equation into a more manageable form.