Problem 23 - Intermediate Algebra

Question

Solve each equation. $$ \frac{3}{x}+\frac{7}{x-1}=1 $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{11 + \sqrt{109}}{2}\) and \(x = \frac{11 - \sqrt{109}}{2}\).

Step 1: Identify a suitable common denominator

To solve the equation, identify the least common denominator (LCD) of the fractions \(\frac{3}{x}\) and \(\frac{7}{x-1}\). The LCD is \(x(x-1)\).

Step 2: Clear the fractions

Multiply every term in the equation by the common denominator \(x(x-1)\) to eliminate the fractions. This gives:\[3(x-1) + 7x = x(x-1) \]

Step 3: Simplify the equation

Expand the equation by multiplying out the terms:\[3x - 3 + 7x = x^2 - x \]This simplifies to:\[ 10x - 3 = x^2 - x \].

Step 4: Move all terms to one side

Rearrange the equation to set it to zero:\[ x^2 - 11x + 3 = 0 \].

Step 5: Use the quadratic formula

Since the equation \(x^2 - 11x + 3 = 0\) is a quadratic equation, use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-11\), and \(c=3\).

Step 6: Calculate the discriminant

Calculate \(b^2 - 4ac\) as follows:\[b^2 - 4ac = (-11)^2 - 4(1)(3) = 121 - 12 = 109\].

Step 7: Substitute in quadratic formula

Substitute the values into the quadratic formula:\[x = \frac{11 \pm \sqrt{109}}{2}\]. This results in two potential solutions for \(x\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). To find the solutions for \( x \), we can use the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a \), \( b \), and \( c \) are coefficients from the equation. This formula helps to find the values of \( x \) that satisfy the equation. When you apply it, you calculate two solutions using the \( \pm \) sign, representing the plus and minus operations. Thus, you get up to two potential solutions for the quadratic equation.
Because we used it here in our original problem, even if the equation appears complex at first, we were able to identify exact values quickly.

Common Denominator

Finding a common denominator is an essential step when dealing with equations that have fractions. The least common denominator (LCD) is the smallest multiple that all denominators can divide into evenly. This allows you to combine or eliminate fractions in an equation.
In the example \( \frac{3}{x} + \frac{7}{x-1} = 1 \), the denominators are \( x \) and \( x-1 \). To find the LCD, multiply these denominators together, resulting in \( x(x-1) \). This makes it possible to clear the fractions from the equation by multiplying each term by this common denominator. Using a common denominator simplifies complex fractions and changes the equation into a form that is easier to solve.

Fraction Elimination

Fraction elimination is the process of removing fractions from an equation to make it easier to work with. Once you identify the common denominator, multiply every term in the equation by it. For example, with the common denominator \( x(x-1) \), this step transforms the original equation:

• \( 3(x-1) + 7x = x(x-1) \)

This procedure effectively ‘clears’ fractions, leaving a simpler polynomial equation that is free of fractional components.
Fraction elimination not only simplifies the visual complexity of an equation but also prepares it for further algebraic steps, such as expanding, factoring, or using the quadratic formula.

Discriminant Calculation

The discriminant is a component within the quadratic formula that provides valuable insights into the nature of the solutions. Calculated as \( b^2 - 4ac \), the discriminant tells you:

• If it is positive, the equation has two real solutions.
• If it is zero, there is exactly one real solution.
• If it is negative, there are two complex solutions.

In our exercise, the discriminant was calculated as 109, which is positive. This indicates that the equation has two distinct real solutions. Knowing the discriminant can speed up the problem-solving process by helping predict what type of solutions you will obtain before completing the quadratic formula. It also reinforces understanding of the types of roots possible for quadratic equations.