Problem 34 - Intermediate Algebra

Question

Solve each proportion. $$ \frac{x-1}{x+1}=\frac{2}{3 x} $$

Step-by-Step Solution

Answer

The solutions are \(x = 2\) and \(x = -\frac{1}{3}\).

Step 1: Cross Multiply

To eliminate the fractions, cross multiply the terms. This means multiplying the numerator of the first fraction by the denominator of the second fraction and vice versa: \((x - 1) \times (3x) = (2) \times (x + 1)\).

Step 2: Distribute and Simplify

Distribute the terms in the cross multiplication. So we have: \(3x(x - 1) = 2(x + 1)\). This simplifies to \(3x^2 - 3x = 2x + 2\).

Step 3: Move All Terms to One Side

To solve for \(x\), move all terms to one side to set up a quadratic equation: \(3x^2 - 3x - 2x - 2 = 0\). This becomes \(3x^2 - 5x - 2 = 0\).

Step 4: Use the Quadratic Formula

Since this is a quadratic equation, use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -5\), and \(c = -2\).

Step 5: Calculate the Discriminant

Calculate \(b^2 - 4ac\): \((-5)^2 - 4 \times 3 \times (-2) = 25 + 24 = 49\).

Step 6: Solve for x

Substitute back into the quadratic formula: \(x = \frac{5 \pm \sqrt{49}}{6}\). Since \(\sqrt{49} = 7\), we have \(x = \frac{5 \pm 7}{6}\).

Step 7: Find the Two Possible Solutions

Calculate both possible values of \(x\): \(x = \frac{12}{6} = 2\) and \(x = \frac{-2}{6} = -\frac{1}{3}\).

Step 8: Verify the Solutions

Substitute \(x = 2\) and \(x = -\frac{1}{3}\) back into the original proportion to check if they satisfy the equation. You will find both solutions work.

Key Concepts

Cross Multiplication Made Simple

Cross multiplication is a handy technique that allows you to solve proportions, which are equations stating that two fractions are equal. It involves multiplying across the equals sign to eliminate the fractions. In other words, you "cross" the terms by multiplying the numerator of one fraction by the denominator of the other fraction and vice versa. This transforms the equation into something that is easier to work with.
Imagine you have the equation \(\frac{x-1}{x+1}=\frac{2}{3x}\). By cross multiplying, you take \((x-1)\times(3x)\) and \((2)\times(x+1)\), which gives you the equation \(3x(x-1)=2(x+1)\).

Cross multiplication is particularly useful in creating a single equation without fractions, making complex expressions much easier to handle. This method is ideal for solving proportion problems since it quickly gets rid of unwieldy fractions and allows you to focus on simpler calculations.

Using the Quadratic Formula

Once you've performed cross multiplication and simplified the equation, you might end up with a quadratic equation. A quadratic equation is typically in the form \(ax^2 + bx + c = 0\). These can be solved using the quadratic formula:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Here, \(a\), \(b\), and \(c\) are coefficients from the quadratic expression. In our exercise, we've rearranged terms from \(3x(x-1) = 2(x+1)\) to form the equation \(3x^2 - 5x - 2 = 0\).
This is where the quadratic formula comes into play. Simply plug in \(a = 3\), \(b = -5\), and \(c = -2\) into the formula and calculate the discriminant \(b^2 - 4ac\). If this value is a perfect square, you can easily find the roots, or solutions, for \(x\). In this case, it turns out to be 49, and since \(\sqrt{49} = 7\), you can find the two potential solutions for \(x\) easily.
The quadratic formula is powerful because it provides a systematic way to find the roots of any quadratic equation, even if the roots are not rational or easy to spot by simpler factorization methods.

Simplifying Expressions Properly

Simplifying expressions is a crucial part of solving algebraic equations, and it comes into play multiple times during the problem-solving process. After cross multiplication, you have the equation \(3x(x-1) = 2(x+1)\). Simplifying involves distributing and combining like terms.
You start by distributing to remove the parentheses: \(3x^2 - 3x = 2x + 2\). Next, you rearrange everything to set the stage for using the quadratic formula, which in this case, results in \(3x^2 - 5x - 2 = 0\).
Efficiency in simplifying expressions can greatly reduce the complexity of calculations. It helps in identifying errors early and ensures that each subsequent step is set up correctly.

• Start by distributing any multipliers across terms in parentheses.
• Combine like terms by adding or subtracting them as needed.
• Move all terms to one side of the equation to prepare for methods like the quadratic formula.

Simplifying expressions effectively paves the way to solving problems more accurately and with greater confidence. Simplification condenses equations into manageable forms, making subsequent calculations straightforward and error-free.