Problem 22 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(\frac{x}{2-x}+\frac{2}{x}=5\)

Step-by-Step Solution

Answer

The solutions are \( x = 1 + \frac{\sqrt{3}}{3} \) and \( x = 1 - \frac{\sqrt{3}}{3} \).

Step 1: Identify a common denominator

First, identify a common denominator for the fractions in the equation \(\frac{x}{2-x}+\frac{2}{x}=5\). The common denominator for \( \frac{x}{2-x} \) and \( \frac{2}{x} \) is \(x(2-x)\).

Step 2: Multiply the entire equation by the common denominator

To eliminate the fractions, multiply every term in the equation by \(x(2-x)\): \[ x(2-x) \times \frac{x}{2-x} + x(2-x) \times \frac{2}{x} = 5x(2-x) \]

Step 3: Simplify the equation

Simplify each term: \[ x^2 + 2(2-x) = 5x(2-x) \] \[ x^2 + 4 - 2x = 10x - 5x^2 \]

Step 4: Combine like terms

Move all terms to one side to form a quadratic equation: \[ x^2 + 4 - 2x - 10x + 5x^2 = 0 \] \[ 6x^2 - 12x + 4 = 0 \]

Step 5: Factor the quadratic equation

Factor out the greatest common factor: \[ 2(3x^2 - 6x + 2) = 0 \] \[ 3x^2 - 6x + 2 = 0 \]

Step 6: Solve the quadratic equation using the quadratic formula

Apply the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \], where \( a = 3\), \( b = -6\), and \( c = 2\): \[ x = \frac{6 \pm \sqrt{36 - 24}}{6} \] \[ x = \frac{6 \pm \sqrt{12}}{6} \] \[ x = \frac{6 \pm 2\sqrt{3}}{6} \] \[ x = 1 \pm \frac{\sqrt{3}}{3} \]

Step 7: Check the solutions in the original equation

Substitute each possible solution back into the original equation to ensure they do not result in undefined fractions: \( x = 1 + \frac{\sqrt{3}}{3} \) and \( x = 1 - \frac{\sqrt{3}}{3} \).

Key Concepts

common denominator

In rational equations, fractions can make solving the problem tricky. To simplify things, find a common denominator. This will allow you to combine the fractions into a single equation without fractions.
In the equation \( \frac{x}{2-x} + \frac{2}{x} = 5 \), the common denominator is \( x(2-x) \). This common denominator helps eliminate fractions, making the equation easier to solve.

quadratic formula

After simplifying the rational equation, we often end up with a quadratic equation. To solve this, we can use the quadratic formula, which is:\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
In our problem, the equation simplifies to \( 3x^2 - 6x + 2 = 0 \). Assign \( a = 3 \), \( b = -6 \), and \( c = 2 \).
By substituting into the formula, we get:\[ x = \frac{6 \pm \sqrt{36 - 24}}{6} \]\[ x = \frac{6 \pm \sqrt{12}}{6} \]\[ x = \frac{6 \pm 2\sqrt{3}}{6} \]\[ x = 1 \pm \frac{\sqrt{3}}{3} \]
This provides the possible solutions for our equation.

simplify equations

Simplifying equations is an essential step in solving more complex problems. Once you have a common denominator, you multiply every term in the equation by this denominator. For instance:\[ x(2-x) \times \frac{x}{2-x} + x(2-x) \times \frac{2}{x} = 5x(2-x) \]
This results in removing the fractions and creating simpler terms to work with. After expanding and combining like terms, you will often end up with a quadratic equation:\[ 6x^2 - 12x + 4 = 0 \].
From here, you can factor or apply the quadratic formula.

check solutions

Always check your solutions to ensure they fit the original equation. This step is crucial for rational equations to avoid undefined values.
Substitute your solutions back into the original equation \( \frac{x}{2-x} + \frac{2}{x} = 5 \).
For \( x = 1 + \frac{\sqrt{3}}{3} \) and \( x = 1 - \frac{\sqrt{3}}{3} \), ensure neither result in a zero denominator. Checking will confirm the solutions are correct and valid for the given problem.