Problem 15 - Algebra for College Students

Question

Find the solution set to each equation. $$\frac{100}{x}=\frac{150}{x+5}-1$$

Step-by-Step Solution

Answer

The solution set is \( x = 43.5 \) or \( x = 11.5 \).

Step 1: Clear the fractions

Multiply through by the common denominator, which is \( x(x+5) \), to eliminate the fractions:\[ 100(x + 5) = 150x - x(x + 5) \]

Step 2: Simplify and rearrange the equation

Expand and simplify the equation from Step 1:\[ 100x + 500 = 150x - x^2 - 5x \]Combine like terms:\[ x^2 - 55x + 500 = 0 \]

Step 3: Solve the quadratic equation

Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -55 \), and \( c = 500 \):\[ x = \frac{55 \pm \sqrt{3025 - 2000}}{2} = \frac{55 \pm \sqrt{1025}}{2} \]

Step 4: Simplify the solutions

Calculate the square root of 1025 and simplify the results:\[ x = \frac{55 \pm 32}{2} \]Thus, \( x = 43.5 \) or \( x = 11.5 \)

Key Concepts

quadratic formula

The quadratic formula is a key tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It states that the solutions for \(x\) are: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] We use this formula when the equation is in the standard quadratic form. Here, \(a\) is the coefficient of \(x^2\), b is the coefficient of \(x\), and \(c\) is the constant term. In our example, once we have the equation simplified to \( x^2 - 55x + 500 = 0 \), we can identify \(a = 1\), \(b = -55\), and \(c = 500\). Let's plug these values into the formula to get: \[x = \frac{55 \pm \sqrt{55^2 - 4 \cdot 1 \cdot 500}}{2 \cdot 1} = \frac{55 \pm \sqrt{3025 - 2000}}{2} = \frac{55 \pm \sqrt{1025}}{2}\] This will yield two solutions for \(x\).

clearing fractions

Clearing fractions involves eliminating fractions from an equation to make it simpler to solve. To do this, we find a common denominator and multiply every term in the equation by it. For the equation: \(\frac{100}{x} = \frac{150}{x+5} - 1\), the common denominator is \(x(x+5)\). Multiplying through by \(x(x+5)\) clears the fractions: \[100(x + 5) = 150x - x(x + 5)\] This multiplication gives us a new equation without fractions, making the next steps easier.

combining like terms

Combining like terms means collecting terms with the same variable to simplify an equation. In our example, we start with the equation after clearing fractions: 100(x + 5) = 150x - x(x + 5). First, expand the terms: \[100x + 500 = 150x - x^2 - 5x\] Next, move all terms to one side to set the equation to zero: \[-x^2 + 100x + 500 - 150x + 5x = 0\] Combine all terms involving \(x\) and constants into single terms: \[x^2 - 55x + 500 = 0\] Now, we have a simplified quadratic equation.

simplifying equations

Simplifying equations involves reducing them to a form that is easier to solve. After clearing fractions and combining like terms, we get the quadratic equation: \(x^2 - 55x + 500 = 0\). To solve it efficiently, we use the quadratic formula to find the roots. Once we apply the formula and calculate the discriminant \(\sqrt{1025}\), we get: \[ x = \frac{55 \pm 32}{2} = \frac{87}{2}\] and \[ \frac{23}{2}\]. Thus, the solutions are: \( x = 43.5\) and \( x = 11.5\). This is a simpler form showing our equation's roots.