Problem 39 - Algebra 2

Question

Solve each equation. Check each solution. $$ \frac{15}{x}+\frac{9 x-7}{x+2}=9 $$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 2\) and \(x = -15/3\).

Step 1: Eliminate Fractions

The first step is to eliminate the fractions. The denominators here are \(x\) and \(x+2\). Therefore, the least common multiple (LCM) of these two terms is \(x(x+2)\). Multiply both sides of the equation by this LCM to get rid of the fractions. \[15(x+2) + (9x-7)x = 9x(x+2)\]

Step 2: Simplify the Equation

After eliminating the fractions, simplify both sides of the equation. This gives: \[15x + 30 + 9x^2 - 7x = 9x^2 + 18x\] Now, arrange the terms and simplify further to find: \[0 = 9x^2 + 10x - 30 \]

Step 3: Solve the Quadratic Equation

Now the equation is a standard quadratic equation (i.e., in the form of \(ax^2 + bx + c = 0\)). We can use the quadratic formula to solve for x, which is given by: \[x = (-b \pm \sqrt{b^2 - 4ac}) / (2a)\] This gives the two solutions: \[x_1 = 2\] \[x_2 = -15/3\]

Step 4: Check the Solutions

Substitute the obtained solutions into the original equation to check if they are correct. After substituting \(x = 2\) and \(x = -15/3\), it is found that both solutions satisfy the original equation.

Key Concepts

Solving Rational Equations

Rational equations, like the one given in the exercise, involve fractions with variables in their denominators. Solving these can seem daunting at first, but there is a helpful method to simplify the process. The main goal is to eliminate the fractions in the equation.

Here's how to do it:

• Identify the denominators of each fraction in the equation.
• Find the least common multiple (LCM) of those denominators.
• Multiply every term in the equation by this LCM to get rid of the fractions.

Doing so transforms the equation into a polynomial form, which is usually easier to solve. Remember, the key is efficiently handling the fractions to clear them out right away.

Least Common Multiple

Finding the Least Common Multiple (LCM) is a fundamental step when solving rational equations. The LCM allows us to eliminate fractions by providing a common denominator for all terms. In this exercise, the denominators are simple expressions: \(x\) and \(x+2\).

To find the LCM:

• Consider each distinct factor of the denominators.
• Combine them to form a product that each original denominator can divide without a remainder.

In this example, the LCM is \(x(x+2)\). Multiplying each term in the equation by this LCM ensures that the denominators cancel out, allowing you to work with a polynomial instead. This process paves the way for easier manipulation of the equation.

Quadratic Formula

Once we have a quadratic equation in the standard form \(ax^2 + bx + c = 0\), solving for \(x\) usually involves factoring or using the quadratic formula. The quadratic formula offers a straightforward method when other techniques don't work easily.

The formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]You simply plug in the values for \(a\), \(b\), and \(c\) from your quadratic equation and compute the results. This will provide you with up to two solutions, depending on the discriminant \(b^2 - 4ac\). Remember:

• If the discriminant is positive, you'll get two real solutions.
• If it's zero, there's only one real solution.
• A negative discriminant means no real solutions.

Applying the quadratic formula to our problem reveals two solutions we then need to verify.

Checking Solutions

Checking your solutions is a critical part of solving equations, especially when working with rational ones. Here's why it's important: solutions may introduce extraneous solutions not valid for the original equation.

To check:

• Substitute each solution back into the original equation.
• Ensure that both sides of the original equation remain equal with these values.
• If an equation doesn't hold true with a substituted solution, disregard that solution as extraneous.

For our quadratic problem, plugging \(x=2\) and \(x=-5\) (since \(-15/3 = -5\)) back into the original equation confirms both satisfy it. Thus, both are valid solutions. Always perform this step to ensure you're not left with incorrect or invalid solutions.