Problem 119 - College Algebra Essentials

Question

In Exercises \(115-122,\) find all values of \(x\) satisfying the given conditions. $$ y_{1}=\frac{2 x}{x+2}, y_{2}=\frac{3}{x+4}, \text { and } y_{1}+y_{2}=1 $$

Step-by-Step Solution

Answer

The possible values of x that satisfy the given conditions are \(x = \frac{-5 + \sqrt{29}}{2}\) and \(x = \frac{-5 - \sqrt{29}}{2}\).

Step 1: Substitute \(y_{1}\) and \(y_{2}\) in the given equation

Given that \(y_{1} + y_{2} = 1\) we can substitute the values of \(y_{1}\) and \(y_{2}\) from our initial conditions into this equation. After substitution, we get \(\frac{2 x}{x+2} + \frac{3}{x+4} = 1\)

Step 2: Simplify the equation

To make progress in solving for \(x\), we'll want to get rid of the fractions. Multiply each term by \((x+2)(x+4)\) to eliminate the denominators: \( 2x(x + 4) + 3(x + 2) = (x + 2)(x + 4)\). This simplifies to the quadratic equation \(2x^{2} + 8x + 3x + 6 = x^{2} + 6x + 8\).

Step 3: Solve the Quadratic Equation

First, subtract \(x^{2} + 6x + 8\) from both sides to set the equation to zero, we then get \(x^{2} + 5x - 2 = 0\). Now factor the quadratic equation or use the quadratic formula. The quadratic formula being \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\). Substituting \(a = 1, b = 5, c = -2\) into the formula gives: \(x = \frac{-5 \pm \sqrt{(5)^{2} - 4(1)(-2)}}{2(1)} = \frac{-5 \pm \sqrt{29}}{2}\) Hence, \(x\) can be \(\frac{-5 + \sqrt{29}}{2}, \frac{-5 -\sqrt{29}}{2}\).

Key Concepts

Quadratic Formula

The quadratic formula is a key tool in algebra, especially when dealing with equations of the form \(ax^2 + bx + c = 0\). This formula helps you find the values of \(x\) that satisfy the equation. It's written as: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Here's how it works:

• Identify the coefficients \(a\), \(b\), and \(c\) in your quadratic equation.
• Plug these into the formula. The \(\pm\) symbol indicates that you will perform two separate calculations: one with addition and one with subtraction.
• Calculate the discriminant, \(b^2 - 4ac\). It tells you how many solutions you might expect: positive for two real solutions, zero for one real solution, and negative for no real solutions.

In our exercise, substituting \(a = 1\), \(b = 5\), and \(c = -2\) into the formula gave us: \[x = \frac{-5 \pm \sqrt{29}}{2}\] This means the solutions are \(\frac{-5 + \sqrt{29}}{2}\) and \(\frac{-5 - \sqrt{29}}{2}\). The discriminant \(29\) is positive, indicating two distinct real solutions.

Fractions Simplification

Simplifying fractions is a fundamental step in solving algebraic equations, especially when fractions make an equation look daunting. In our problem, we had: \[\frac{2x}{x+2} + \frac{3}{x+4} = 1\] To simplify, it's often helpful to clear the denominators. Here, we multiplied every term by \((x+2)(x+4)\) to eliminate them. This is because:

• It converts the equation into a polynomial form, which is usually easier to manipulate.
• We get rid of the fractions, allowing us to focus on solving the equation directly.

After multiplying, the equation becomes: \[2x(x + 4) + 3(x + 2) = (x + 2)(x + 4)\] This simplification leads us to a quadratic equation, which we can solve using other methods like factoring or the quadratic formula. Remember, always perform the same action (like multiplying or adding) to both sides of the equation to maintain balance.

Solving Quadratic Equations

Solving quadratic equations, either through factoring, completing the square, or using the quadratic formula, is essential in algebra. In our example, after eliminating the fractions and obtaining\[2x^2 + 8x + 3x + 6 = x^2 + 6x + 8\] we simplify to: \[x^2 + 5x - 2 = 0\] This is a standard quadratic equation. Solving it involves either factoring or applying the quadratic formula. Factoring requires finding two numbers that multiply to \(-2\) and add to \(5\), but sometimes that can be challenging. Applying the quadratic formula is straightforward here, allowing us to find: \[x = \frac{-5 \pm \sqrt{29}}{2}\]

• This approach is universal and works even when factoring is complicated or the numbers aren't nice.
• Always check if the equation can be factored to save time, but when in doubt, the quadratic formula is your best friend.

By following these steps, you can solve any quadratic equation, ensuring you find all possible values of \(x\) that satisfy your original mathematical conditions.