Problem 24 - BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Question

Solve the equation by using the LCD. Check your solution(s). $$\frac{10}{x}+3=\frac{x+9}{x-4}$$

Step-by-Step Solution

Answer

The solution to the equation is \(x = -10\).

Step 1: Find the LCD

The left hand side of the equation has denominators \(x\) and the right hand side has denominators \(x-(4)\). The Least Common Denominator (LCD) of these fractions is their product \(x(x-4)\).

Step 2: Simplify the Equation

Multiply each term of the equation by the LCD, \(x(x-4)\), to remove the denominators. This gives: \(10(x-4) + 3x(x-4) = (x+9)x, \)Simplify this into:\(10x - 40 + 3x^2 - 12x = x^2 + 9x, \)Rearrange by deducting both sides by \(x^2 + 9x\):\(3x^2 - x^2 + 10x - 9x - 40 = 0, \)Combining like terms this simplifies into:\(2x^2 + x - 40 = 0.\)

Step 3: Solve x from the Equation

Now, solve \(x\) from the resulting quadratic equation \(2x^2 + x - 40 = 0\) by use of the quadratic formula. This gives two possible results: \(x = \frac{-1 + \sqrt{1^2 - 4.(-40).2}}{2.2} = 4 \)and\(x = \frac{-1 - \sqrt{1^2 - 4.(-40).2}}{2.2} = -10 \)

Step 4: Check the solutions

Solutions should also satisfy the original equation. When \(x=4\) is substituted into the original equation, it gives \(\frac{10}{4} + 3 \neq \frac{4+9}{0}\), hence \(x\) cannot be 4. Substituting \(x=-10\) into the original equation, it gives \(\frac{10}{-10} + 3 = \frac{-10+9}{-10-4}\), so \(x = -10\) is the correct solution.

Key Concepts

Least Common Denominator

When solving equations that include fractions, especially with different denominators, finding the Least Common Denominator (LCD) is crucial. The LCD is the smallest expression that each of the denominators can divide evenly. It allows us to eliminate the fractions by converting them into terms that have a common denominator, making computations much simpler.
For example, consider the equation given:

• The fraction on the left has a denominator of \(x\), and the fraction on the right has a denominator of \(x-4\).
• The LCD for these denominators is their product, \(x(x-4)\).

By multiplying the entire equation by this LCD, we can remove the fractions altogether, allowing for a direct approach to solving the equation.

Quadratic Formula

The quadratic formula is used to find the roots of a quadratic equation, which is an equation of the form \(ax^2 + bx + c = 0\). This powerful tool provides a straightforward method of finding the solutions for \(x\) in such equations.

• The standard quadratic formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
• In our context, the quadratic equation derived was \(2x^2 + x - 40 = 0\), where \(a = 2\), \(b = 1\), and \(c = -40\).
• Plugging these values into the quadratic formula enabled finding two potential solutions for \(x\).

The solutions are the \(x\) values that make the equation zero, but we must verify if they satisfy the original equation.

Checking Solutions

Once solutions are found using methods such as the quadratic formula, it is important to verify their validity by substituting them back into the original equation. This step ensures the solutions don't make any part of the equation undefined or result in any mathematical contradictions.
In the example:

• For \(x = 4\), substituting back into the original equation resulted in division by zero. Hence, it is invalid.
• For \(x = -10\), substituting it back yielded a true statement, confirming it as a valid solution.

This checking process is essential to avoid errors, particularly when dealing with rational equations that can have restrictions on solutions.

Algebra

Algebra is the branch of mathematics dealing with symbols and the rules for manipulating these symbols. It involves operations with variables, constants, and expressions, leading to equations to be solved. This particular problem demonstrates several key algebraic processes:

• Identifying common denominators to eliminate fractions.
• Reorganizing and simplifying terms to form a quadratic equation.
• Applying the distributive property, combining like terms, and isolating variable terms to solve equations.

These steps reflect fundamental algebraic principles that are necessary for solving complex equations effectively. Understanding these core concepts in algebra is important for helping students maneuver through similar mathematical challenges with confidence.