Problem 16 - Algebra 2 and Trigonometry

Question

Solve each proportion for the variable. \(\frac{2}{x-1}=\frac{x+2}{2}\)

Step-by-Step Solution

Answer

The solutions are \( x = -3 \) and \( x = 2 \).

Step 1: Cross Multiply

To solve the proportion \(\frac{2}{x-1}=\frac{x+2}{2}\), we use the method of cross multiplication. This means multiplying the numerator of the first ratio by the denominator of the second ratio and setting it equal to the product of the denominator of the first ratio and the numerator of the second ratio: \[ 2 \cdot 2 = (x - 1)(x + 2) \] This simplifies to: \[ 4 = (x - 1)(x + 2) \]

Step 2: Expand and Simplify

Now, expand the expression on the right side of the equation:\[ 4 = (x-1)(x+2) \]Use the distributive property (FOIL) to expand the equation:\[ 4 = x^2 + 2x - x - 2 \] Simplify the equation:\[ 4 = x^2 + x - 2 \]

Step 3: Move All Terms to One Side

To solve the quadratic equation, move all terms to one side to set the equation to zero:\[ x^2 + x - 2 - 4 = 0 \] Simplify:\[ x^2 + x - 6 = 0 \]

Step 4: Factor the Quadratic Equation

Next, factor the quadratic equation \( x^2 + x - 6 = 0 \). Look for two numbers that multiply to -6 and add to 1. These numbers are 3 and -2. Write the equation as:\[ (x + 3)(x - 2) = 0 \]

Step 5: Solve for the Variable

We have factored the equation as \( (x + 3)(x - 2) = 0 \). Set each factor equal to zero to solve for \(x\):1. \( x + 3 = 0 \) which gives \( x = -3 \)2. \( x - 2 = 0 \) which gives \( x = 2 \)Thus, the solutions are \( x = -3 \) and \( x = 2 \).

Step 6: Verify the Solutions

Verify both solutions by substituting back into the original equation, \( \frac{2}{x-1}=\frac{x+2}{2} \):- Checking \( x = -3 \): LHS: \( \frac{2}{-3-1} = \frac{2}{-4} = -\frac{1}{2} \) RHS: \( \frac{-3+2}{2} = \frac{-1}{2} = -\frac{1}{2} \) Since LHS = RHS, \( x = -3 \) is a valid solution.- Checking \( x = 2 \): LHS: \( \frac{2}{2-1} = \frac{2}{1} = 2 \) RHS: \( \frac{2+2}{2} = \frac{4}{2} = 2 \) Since LHS = RHS, \( x = 2 \) is a valid solution.Both solutions satisfy the original equation.

Key Concepts

Cross Multiplication

Cross multiplication is an essential tool when solving proportions. A proportion is an equation that states two ratios are equal. In this exercise, we started with the proportion \( \frac{2}{x-1} = \frac{x+2}{2} \). Cross multiplication involves taking each numerator and multiplying it by the opposite denominator. This allows us to move from a fraction to a straightforward algebraic equation. Here’s how it works: Multiply 2 (the numerator of the first fraction) by 2 (the denominator of the second fraction) and set it equal to \((x-1)\) (the denominator of the first fraction) times \((x+2)\) (the numerator of the second fraction). The result is \(4 = (x - 1)(x + 2)\). This transforms the proportion into an equation we can solve using algebraic methods.

Factoring Quadratic Equations

After cross-multiplication, our equation is \(4 = x^2 + x - 2\). Our next task is factoring a quadratic equation. A quadratic equation usually takes the form \(ax^2 + bx + c = 0\). Here, it stands as \(x^2 + x - 6 = 0\) after simplifying. Factoring helps us break this equation into simpler pieces called factors. We need two numbers that multiply to \(-6\) and add to \(1\). The numbers here are \(3\) and \(-2\). This gives us the factors \((x + 3)\) and \((x - 2)\). Therefore, we rewrite our equation as \((x + 3)(x - 2) = 0\). Factoring is crucial because it gives us potential solutions to the equation.

Verifying Solutions

After factoring, we solve for \(x\). The equation \((x + 3)(x - 2) = 0\) offers solutions by setting each factor equal to zero. Solving gives \(x = -3\) and \(x = 2\). Verification is a necessary step to ensure these solutions satisfy the original equation.
Substitute \(x = -3\) into \(\frac{2}{x-1}=\frac{x+2}{2}\) to check:

• Left-hand side (LHS): \( \frac{2}{-3-1} = \frac{2}{-4} = -\frac{1}{2} \)
• Right-hand side (RHS): \( \frac{-3+2}{2} = -\frac{1}{2} \)

LHS matches RHS, so \(x = -3\) works.
For \(x = 2\):

• LHS: \( \frac{2}{2-1} = 2 \)
• RHS: \( \frac{2+2}{2} = 2 \)

Again, LHS equals RHS, showing \(x = 2\) is also valid. Verification assures us both solutions are correct.

Step by Step Solution

Breaking down complex problems into clear, manageable steps is beneficial for solving equations. With a step-by-step approach, each section of the problem becomes easier to follow:

• First, identify the problem structure, such as proportions, encouraging the use of cross multiplication.
• Then, simplify and rearrange the equation, preparing it for factoring by setting it equal to zero.
• Next, factor the quadratic equation to discover potential solutions.
• Finally, verify each solution to ensure accuracy.

This structured method not only guides through the equation but enhances understanding and problem-solving skills. Learning to depend on such a systematic approach is a foundation for tackling increasingly complex algebraic concepts.