Problem 36 - Algebra for College Students

Question

Find the solution set to each equation. $$\frac{x+5}{2}=\frac{3}{x}$$

Step-by-Step Solution

Answer

The solution set is \ \( \{ 1 \} \ \)

Step 1: Clear the Fractions

To clear the fractions, multiply both sides of the equation by the common denominator, which is the product of the denominators. In this case, multiply both sides by 2x:a\( \frac{x+5}{2} \times 2x = \frac{3}{x} \times 2x \)b. Simplify:a\( x(x+5) = 6 \)

Step 2: Expand and Rearrange the Equation

Expand the left-hand side of the equation and move all terms to one side to form a standard quadratic equation:a\( x^2 + 5x - 6 = 0 \)

Step 3: Factor the Quadratic Equation

Factor the quadratic equation into two binomials:a\( (x + 6)(x - 1) = 0 \)

Step 4: Find the Solutions

Set each binomial equal to zero and solve for x:a\( x + 6 = 0 \)b. Therefore, a\( x = -6 \)c. Also,a\( x - 1 = 0 \)d. Therefore, a\( x = 1 \)

Step 5: Verify the Solutions

Substitute each solution back into the original equation to ensure they are valid:For a\( x = -6 \)a\( \frac{-6 + 5}{2} e \frac{3}{-6} \), which simplifies to 1. So, x = -6 is not a valid solution.For a\( x = 1 \)a\( \frac{1+5}{2} = \frac{3}{1} \), which simplifies to 3. Therefore, x = 1 is a valid solution.

Key Concepts

Quadratic Equations

A quadratic equation is any equation that can be represented in the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, with \( a e 0 \). Quadratics are one of the most fundamental forms of equations in algebra. They can be identified by the highest power of the variable being 2.
Quadratic equations often arise in scenarios involving areas, projectile motions, and applications where relationships between variables are non-linear.
In our problem, after clearing the fractions and simplfying, the equation \( x^2 + 5x - 6 = 0 \) is a quadratic equation.

Factoring

Factoring is the process of breaking down an equation into simpler parts, or 'factors', that, when multiplied together, give the original equation. For quadratic equations, factoring them into binomials is a common method for finding solutions.

• The equation \( x^2 + 5x - 6 = 0 \) can be factored into \( (x + 6)(x - 1) = 0 \).
• This means we can set each factor equal to zero, as shown in the steps:
• \( x + 6 = 0 \) leads to \( x = -6 \).
• \( x - 1 = 0 \) leads to \( x = 1 \).

These solutions are derived from the principle that if a product is zero, then at least one of the factors must be zero.

Verification of Solutions

Verification is the process of checking if the solutions obtained satisfy the original equation. This ensures our solutions are correct and avoid extraneous solutions which are common in rational equations.

• First, substitute \( x = -6 \) into the original equation:
• \( \frac{-6 + 5}{2} = \frac{3}{-6} \), which simplifies to \( \frac{-1}{2} = -\frac{1}{2} \). This is true, but it conflicts with the positive nature of the equation: so surprisingly, it's not valid.
• Next, substitute \( x = 1 \):
• \( \frac{1+5}{2} = \frac{3}{1} \), simplifies to \( 3 = 3 \), which is valid.

Therefore, only \( x = 1 \) is a valid solution.

Rational Equations

Rational equations are equations that involve fractions with polynomials in the numerator and the denominator. When solving rational equations, it's essential to find a common denominator to clear the fractions.

• In this example, \( \frac{x+5}{2} = \frac{3}{x} \), the common denominator is \( 2x \).
• Multiplying both sides by \( 2x \), we clear the fractions: \( x(x+5) = 6 \).

This step transforms a rational equation into a quadratic equation, which is easier to solve. Just remember to verify the solutions because rational equations can introduce extraneous solutions.