Problem 54 - Precalculus

Question

Solve each equation for \(x\) in terms of the other letters. $$\frac{1}{a x}=\frac{1}{b x}-\frac{1}{c}$$

Step-by-Step Solution

Answer

Use the quadratic formula after simplifying: \(x = \frac{-bc \pm \sqrt{b^2c^2 + 4a^2bc}}{2ab}\).

Step 1: Get a common denominator on the right-hand side

We start by finding a common denominator for the right-hand side of the equation: \( \frac{1}{b x} - \frac{1}{c} \). The common denominator is \(bxc\). This makes the equation: \( \frac{c}{bcx} - \frac{bx}{bcx} \).

Step 2: Combine the fractions on the right-hand side

Combine the fractions on the right-hand side: \(\frac{c}{bcx} - \frac{bx}{bcx} = \frac{c - bx}{bcx} \). Now the equation is \(\frac{1}{a x} = \frac{c - bx}{bcx}\).

Step 3: Cross-multiply to eliminate fractions

Cross-multiply to get rid of the fractions: \(bcx = ax(c - bx)\).

Step 4: Distribute on the right-hand side

Distribute \(ax\) on the right-hand side of the equation: \(bcx = axc - abx^2\).

Step 5: Rearrange the equation

Rearrange the equation so all terms involving \(x\) are on one side: \( abx^2 + bcx - axc = 0 \).

Step 6: Use the quadratic formula

This is a quadratic equation in the form \(Ax^2 + Bx + C = 0\), where \(A = ab\), \(B = bc\), and \(C = -ac\). Use the quadratic formula: \(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\). Substitute \(A = ab\), \(B = bc\), \(C = -ac\) into the formula.

Step 7: Substitute values into the quadratic formula

Substitute into the quadratic formula: \(x = \frac{-bc \pm \sqrt{(bc)^2 - 4(ab)(-ac)}}{2ab}\). Simplify this expression to get the value of \(x\) in terms of the other letters.

Key Concepts

Cross-Multiplication

Cross-multiplication is a powerful technique often used to solve equations that involve fractions. When you have an equation with two fractions set equal to each other, you can use cross-multiplication to simplify the equation and remove the fractions. This is especially helpful when dealing with algebraic expressions where solving for a variable is necessary.
For instance, if you have an equation in the form \(\frac{a}{b} = \frac{c}{d}\), you can cross-multiply to get \(a \times d = b \times c\).
In the original exercise, after obtaining a single fraction on each side of the equation, we cross-multiplied to eliminate the fractions completely. This step is crucial because it simplifies the equation from a complex fraction form to a more manageable algebraic expression.

Fractions in Equations

Solving equations with fractions involves understanding how to manipulate and simplify fractions to get a clear solution. Fractions can make equations seem complicated, but with the right approach, they can become much simpler. Let's break down the steps:

• Identify a common denominator: Look for a common denominator to combine fractions on one side of the equation. This helps when dealing with multiple fractions as it simplifies the operation.
• Combine fractions: Use the common denominator to rewrite and combine the fractions into a single fraction. This reduces the complexity of the equation.
• Eliminate the fraction: Often, you can multiply both sides by the denominator to get rid of the fraction, making the equation a ‘normal’ equation.

In the original problem, the fractions on the right-hand side were combined into a single fraction, making it easier to perform the cross-multiplication. This is a perfect example of dealing properly with fractions in equations.

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations. A quadratic equation is typically in the form \(Ax^2 + Bx + C = 0\). Solving such an equation involves finding the value(s) of \(x\) that make the equation true.
The quadratic formula is:
x = \( \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \)
Here's how to use it:

• First, identify coefficients \(A\), \(B\), and \(C\) from the equation. Make sure the equation is in the standard quadratic form.
• Substitute these values into the quadratic formula.
• Solve for \(x\) using calculation for the square root and the division.

In the provided exercise, the quadratic formula was used after rearranging the equation into the standard quadratic form. By substituting \(A=ab\), \(B=bc\), and \(C=-ac\) into the formula, one can find the solutions for \(x\). This illustrates how versatile and essential the quadratic formula is for solving quadratic equations.