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Magazine Chapter 11: Problem 16
Solve each of the following exercises algebraically. The numerator of a fraction is 1 less than the denominator. If \(\frac{7}{12}\) is added to the fraction, the result is the reciprocal of the original fraction. Find the original fraction.
Short Answer
Expert verified
The original fraction is \( \frac{3}{4} \).
Step by step solution
01
Define Variables
Let the denominator of the original fraction be represented by the variable \( x \). Since the numerator is 1 less than the denominator, the numerator can be represented as \( x - 1 \). Therefore, the original fraction is \( \frac{x-1}{x} \).
02
Express the Given Condition
According to the problem, when \( \frac{7}{12} \) is added to the original fraction, the result is the reciprocal of the original fraction. This can be written algebraically as: \[ \frac{x-1}{x} + \frac{7}{12} = \frac{x}{x-1} \]
03
Find a Common Denominator
To add the fractions on the left side of the equation, find a common denominator. The common denominator for \( x \) and 12 is \( 12x \). Transform the equation to: \[ \frac{12(x-1) + 7x}{12x} = \frac{x}{x-1} \]
04
Simplify the Equation
Simplify the numerator on the left side: \[ \frac{12x - 12 + 7x}{12x} = \frac{19x - 12}{12x} \] This results in the equation: \[ \frac{19x - 12}{12x} = \frac{x}{x-1} \]
05
Cross Multiply to Solve
To eliminate the fractions, cross-multiply both sides: \[ (19x - 12)(x - 1) = 12x^2 \] Expand and simplify the equation: \[ 19x^2 - 19x - 12x + 12 = 12x^2 \] Combine like terms: \[ 19x^2 - 31x + 12 = 12x^2 \] Move all terms to one side to form a standard quadratic equation: \[ 7x^2 - 31x + 12 = 0 \]
06
Solve the Quadratic Equation
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 7 \), \( b = -31 \), and \( c = 12 \). Plug these values in: \[ x = \frac{31 \pm \sqrt{31^2 - 4 \cdot 7 \cdot 12}}{2 \cdot 7} \] Calculate the discriminant: \[ 31^2 - 4 \cdot 7 \cdot 12 = 961 - 336 = 625 \] Substitute back into the formula: \[ x = \frac{31 \pm 25}{14} \] Therefore, the solutions are: \[ x = \frac{31 + 25}{14} = 4 \] \[ x = \frac{31 - 25}{14} = \frac{6}{14} = \frac{3}{7} \]
07
Verify the Solutions
Check whether each solution makes sense for the fraction. \( x = 4 \) gives the original fraction \( \frac{4-1}{4} = \frac{3}{4} \). \( \frac{3}{7} \) is invalid because the numerator would be negative (\( \frac{\frac{3}{7} - 1}{\frac{3}{7}} \) is undefined for our context). Thus, \( x = 4 \) is the valid solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fractions
Fractions are numerical quantities that represent a part of a whole. A fraction consists of a numerator (the top number) and a denominator (the bottom number). For example, in the fraction \(\frac{3}{4}\), 3 is the numerator and 4 is the denominator. Understanding fractions is crucial because they appear in various mathematical problems, including algebra word problems like the one we are solving here.
- The challenge often lies in manipulating fractions, especially when adding, subtracting or finding common denominators.
- To add fractions, we first need a common denominator. For instance, to add \(\frac{x-1}{x}\) and \(\frac{7}{12}\), we find the common denominator, which is \(12x\).
- Once the denominators are the same, we convert the fractions and then combine the numerators.
Quadratic Equations
Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients. These equations often appear in algebra problems when we are solving for unknown variables.
- In our exercise, we reach a quadratic equation after simplifying the initial algebraic conditions.
- The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), is a powerful tool for solving these equations.
- The discriminant, given by \(b^2 - 4ac\), helps determine the nature of the roots (real and distinct, real and repeated, or complex).
Reciprocals
Reciprocals are numbers that, when multiplied together, equal 1. The reciprocal of a fraction \(\frac{a}{b}\) is \(\frac{b}{a}\). Recognizing and working with reciprocals is essential in many algebraic operations.
- In the exercise, when we are told that the sum of \(\frac{7}{12}\) and the original fraction equals its reciprocal, we need to understand what that implies algebraically.
- Given the original fraction is \(\frac{x-1}{x}\), its reciprocal is \(\frac{x}{x-1}\).
- Setting up the equation correctly ensures that we can solve for \(x\).
Cross-Multiplication
Cross-multiplication is a method used to solve equations involving two fractions set equal to each other. It simplifies comparisons and solutions of fractional equations.
- For the equation \(\frac{19x - 12}{12x} = \frac{x}{x-1}\), we use cross-multiplication to eliminate the fractions.
- To cross-multiply, we multiply the numerator of one fraction by the denominator of the other fraction and set the products equal to each other: \((19x - 12)(x - 1) = 12x^2\).
- This results in a polynomial equation that can be simplified and solved like any other quadratic equation.
