Problem 19 - College Algebra with Modeling and Visualization

Question

Find all real solutions. Check your results. $$ \frac{x+5}{x+2}=\frac{x-4}{x-10} $$

Step-by-Step Solution

Answer

The real solution is \(x = -14\).

Step 1: Cross-Multiply the Fractions

To eliminate the fractions and solve the equation, cross-multiply: \[(x + 5)(x - 10) = (x - 4)(x + 2)\] This step will give us an equation without fractions.

Step 2: Expand Both Sides

Expand both sides of the equation using the distributive property:\[x^2 - 10x + 5x - 50 = x^2 + 2x - 4x - 8\] This simplifies to: \[x^2 - 5x - 50 = x^2 - 2x - 8\]

Step 3: Eliminate Common Terms

Subtract \(x^2\) from both sides to eliminate the quadratic term:\[-5x - 50 = -2x - 8\]

Step 4: Solve for x

Rearrange the equation to solve for \(x\). Start by getting all terms involving \(x\) on one side:\[-5x + 2x = -8 + 50\]Simplify:\[-3x = 42\]Solve for \(x\) by dividing both sides by -3:\[x = -14\]

Step 5: Check the Solution

Substitute \(x = -14\) back into the original fractions to verify:Left side: \(\frac{-14 + 5}{-14 + 2} = \frac{-9}{-12} = \frac{3}{4}\)Right side: \(\frac{-14 - 4}{-14 - 10} = \frac{-18}{-24} = \frac{3}{4}\).Both sides are equal, confirming that \(x = -14\) is a valid solution.

Key Concepts

Cross-Multiplication

Cross-multiplication is a handy technique for solving equations involving fractions. It helps us to eliminate the denominators and work with simpler linear or quadratic equations. Here's how it works: when we have an equation of the form \( \frac{a}{b} = \frac{c}{d} \), we can multiply across the diagonal—hence the term cross-multiplication. This results in the equation \( a \cdot d = b \cdot c \). It is important because it keeps the equation balanced while removing the fractions.
In our problem, we apply this method to \( \frac{x+5}{x+2} = \frac{x-4}{x-10} \). Cross-multiplying gives \((x+5)(x-10) = (x-4)(x+2)\). Now the equation no longer contains fractions, paving the way for further simplification using algebraic rules.

• Ensure equations are set up, so fractions are equal to each other before crossing.
• Do not forget to parenthesize expressions when needed, as in \((x+5)(x-10)\).

Distributive Property

The distributive property is a crucial algebraic rule used to expand expressions, especially when multiplying a binomial by another binomial, as seen in our exercise. This property states that for any three numbers, \( a \), \( b \), and \( c \), \( a(b+c) = ab + ac \). When dealing with our cross-multiplied equation \((x+5)(x-10) = (x-4)(x+2)\), we apply the distributive property to both sides to expand them.
Expanding involves:

• Multiplying each term in the first binomial by each term in the second binomial.
• Carefully keeping track of positive and negative signs.

The result of applying this property allows us to write both sides of the equation without parentheses: \(x^2 - 10x + 5x - 50\) becomes \(x^2 - 5x - 50\), and \(x^2 + 2x - 4x - 8\) simplifies to \(x^2 - 2x - 8\). It is essential to handle these steps cleanly to avoid errors further in the solving process.

Checking Solutions

Once a solution is found, it is essential to check it within the context of the original equation to ensure validity. This step confirms that no mistakes were made during solving, such as arithmetic errors or violations of the original equation's constraints. You substitute the solution back into the original equation and verify that both sides are equal.
In our exercise, substituting \( x = -14 \) back into the original fractions gives us \( \frac{-14 + 5}{-14 + 2} = \frac{3}{4} \) on the left side and \( \frac{-14 - 4}{-14 - 10} = \frac{3}{4} \) on the right side. Since both sides are indeed equal, \( x = -14 \) is a verified solution.

• Use caution to revert back to the original equation, not any manipulated versions.
• Pay special attention to the denominators to ensure no division by zero.

This practice not only confirms correctness but also builds confidence in the steps you've taken to solve algebraic equations.