Problem 16 - Algebra for College Students

Question

Find the solution set to each equation. $$\frac{30}{x}=\frac{50}{x+10}+\frac{1}{2}$$

Step-by-Step Solution

Answer

The solution set is \(x = -60\).

Step 1: Identify a common denominator

The denominators in the equation are: \(x\), \(x+10\), and 2. To eliminate the fractions, identify the common denominator: \(2x(x+10)\).

Step 2: Multiply through by the common denominator

Multiply every term by \(2x(x+10)\) to clear the fractions: \(2x(x+10) \times \frac{30}{x} = 2x(x+10) \times \frac{50}{x+10} + 2x(x+10) \times \frac{1}{2}\). Simplify each term: \(60(x+10) = 100x + x(x+10)\).

Step 3: Expand and Simplify the Equation

Expand and combine like terms: \(60x + 600 = 100x + x^2 + 10x\). Combine like terms: \(60x + 600 = x^2 + 110x\).

Step 4: Rearrange into standard quadratic form

Rearrange all terms to one side to set the equation to zero: \(x^2 + 110x - 60x - 600 = 0\). Simplify: \(x^2 + 50x - 600 = 0\).

Step 5: Solve the quadratic equation

Using the quadratic formula \(x =\frac{-b\pm\sqrt{b^2-4ac}}{2a}\), let \(a=1\), \(b=50\), \(c=-600\): \[ x = \frac{-50\pm\sqrt{50^2-4(1)(-600)}}{2(1)} = \frac{-50\pm\sqrt{2500+2400}}{2} = \frac{-50\pm\sqrt{4900}}{2} = \frac{-50\pm 70}{2}.\] This results in two values: \(x = 10\) and \(x = -60\).

Step 6: Check the solutions

Substitute \(x = 10\) and \(x = -60\) back into the original equation to check for validity. When \(x = 10\), the original equation results in a division by zero, which is undefined, thus, \(x = 10\) is not a valid solution. \(x = -60\) does not cause division by zero and satisfies the equation. Thus, \(x = -60\) is the valid solution.

Key Concepts

quadratic formula

The quadratic formula is a crucial tool used to solve equations of the form \(ax^2 + bx + c = 0\). It provides the solutions by using the coefficients \(a\), \(b\), and \(c\) directly in the formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In this structure:

• \(a\) is the coefficient of \(x^2\),
• \(b\) is the coefficient of \(x\),
• \(c\) is the constant term.

The plus-minus symbol (\(\pm\)) indicates that there are potentially two solutions, one for addition and one for subtraction.
In our example, after rearranging the equation \( \frac{30}{x} = \frac{50}{x+10} + \frac{1}{2} \) and simplifying, we get a quadratic equation: \(x^2 + 50x - 600 = 0\). By plugging \(a = 1\), \(b = 50\), and \(c = -600\) into the quadratic formula, we solve the quadratic equation.

common denominators

To solve rational equations involving fractions, it is essential to find a common denominator. This step harmonizes the fractions, making it easier to eliminate them. For the equation \( \frac{30}{x} = \frac{50}{x+10} + \frac{1}{2} \), the denominators are \(x\), \(x+10\), and 2. Their common denominator is \(2x(x+10)\).
Identifying and using this common denominator allows us to clear the fractions by multiplying every term. This transforms the fractions into simpler polynomial form, making the equation easier to manage.
After multiplying through by \(2x(x+10)\), each term in our example becomes:

• \(60(x+10)\),
• \(100x\),
• and \(x(x+10)\).

This step sets the stage for further simplification and solving the equation.

fraction elimination

Fraction elimination is a critical step in solving rational equations. By multiplying each part of the equation by a common denominator, you eliminate the fractions, simplifying the equation to a more workable form.
In our example, after multiplying both sides by the common denominator \(2x(x+10)\), the equation becomes:

• \(60(x+10) = 100x + x(x+10)\)

By expanding and combining like terms, the equation simplifies to a quadratic form.
This method helps in transforming a complex fractional equation into a simpler polynomial equation, which can be tackled using algebraic techniques such as factoring or the quadratic formula.

validating solutions

After solving a rational equation, it's crucial to check the validity of the solutions by substituting them back into the original equation. This step ensures that the solutions are correct and do not introduce any undefined conditions, such as division by zero.
In our example, the solutions obtained were \(x = 10\) and \(x = -60\).

• Substituting \(x = 10\) results in division by zero, making it invalid.
• Substituting \(x = -60\) does not cause any undefined conditions and satisfies the equation, thus it is a valid solution.

Always remember, validating solutions is vital in identifying extraneous solutions that do not fit the original conditions of the equation. This final step ensures your solution set is accurate and reliable.