/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 $$ \frac{2}{x-14}-\frac{5}{x-1... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 51

$$ \frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11} $$

Short Answer

Expert verified
The short answer version: To solve the equation \(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}\), we first find the least common denominator (LCD) which is (x-14)(x-13)(x-9)(x-11). Then, we multiply both sides of the equation by the LCD to eliminate fractions. After simplification and solving for x, we get the quadratic equation \((x-11)(x-13) = (x-14)(x-9)\). Solving this equation gives us x=17 as the solution. We then check for extraneous solutions, and since x=17 doesn't make any denominator zero, it is a valid solution. Thus, the answer is \(x=17\).

Step by step solution

01

Find the least common denominator (LCD)

To find the least common denominator of the fractions, we will need to find the Lowest Common Multiple (LCM) of the four denominators: (x-14), (x-13), (x-9), and (x-11). Since the denominators are all different, their LCM is simply their product: LCD = (x-14)(x-13)(x-9)(x-11)
02

Multiply each term by the LCD

Now, we will multiply each term by the LCD to eliminate the denominators: \((x-14)(x-13)(x-9)(x-11)\left(\frac{2}{x-14}\right) - (x-14)(x-13)(x-9)(x-11)\left(\frac{5}{x-13}\right) = (x-14)(x-13)(x-9)(x-11)\left(\frac{2}{x-9}\right) - (x-14)(x-13)(x-9)(x-11)\left(\frac{5}{x-11}\right)\) Now, simplify the equation: \((x-13)(x-9)(x-11)(2) - (x-14)(x-9)(x-11)(5) = (x-14)(x-13)(x-11)(2) - (x-14)(x-13)(x-9)(5)\)
03

Solve the equation for x

Now, we will solve the equation for x by expanding the equation, collecting the like terms, and solving for x: \(2(x-11)(x-9)(x-13) - 5(x-14)(x-11)(x-9) = 2(x-14)(x-11)(x-13) - 5(x-14)(x-9)(x-13)\) Now let's move the terms from the right side to the left to have all the terms on one side: \( 2(x-11)(x-9)(x-13) - 5(x-14)(x-11)(x-9) - 2(x-14)(x-11)(x-13) + 5(x-14)(x-9)(x-13) = 0 \) Now factor the common terms: \( (x-11)(x-13)[2(x-9) - 5(x-14)] - (x-14)(x-9)[2(x-11) - 5(x-13)] = 0 \) Now simplify: \( (x-11)(x-13)(2x - 18 - 5x + 70) - (x-14)(x-9)(2x - 22 - 5x + 65) = 0 \) \( (x-11)(x-13)(-3x + 52) - (x-14)(x-9)(-3x + 43) = 0 \) Let \(u = -3x\), so the equation becomes: \( (x-11)(x-13)(u+52) - (x-14)(x-9)(u+43) = 0 \) Now, we're assuming that there is a unique solution. If there isn't, this won't work: \((x-11)(x-13) = (x-14)(x-9)\) This is a quadratic equation. Expand and simplify: \(x^2 - 24x + 143 = x^2 - 23x + 126\) Simplify further: \(-x + 17 = 0\) Hence, x = 17.
04

Check for extraneous solutions

Check if the solution of x = 17 is not an extraneous solution, which means it doesn't make any of the original denominators zero. For our given equation: \(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}\) We can check that none of the denominators become zero when x=17: x-14 = 3 x-13 = 4 x-9 = 8 x-11 = 6 Since the denominators do not become zero, x=17 is a valid solution. Therefore, our final answer is: \(x = 17\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Least Common Denominator
Understanding the least common denominator (LCD) is vital when working with algebraic fractions. The LCD is the smallest expression that all denominators in a given set of fractions can divide into without leaving a remainder. To find the LCD, we must identify the least common multiple (LCM) of the denominators. For distinct linear factors, such as those often found in algebraic equations, the LCD is simply the product of these factors.

For instance, in the exercise \(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}\), we look for a common denominator between \(x-14\), \(x-13\), \(x-9\), and \(x-11\). Since these are all different, the LCD is their product. This step enables us to combine the fractions and work with a unified denominator, paving the way for simpler algebraic manipulations.
Solving Quadratic Equations
A quadratic equation is a second-degree polynomial equation, usually in the form \(ax^2+bx+c=0\). Solving a quadratic equation might involve factoring, completing the square, using the quadratic formula, or graphing, depending on the structure of the equation.

The process begins with setting the equation to zero and finding factors that produce the original quadratic when multiplied. Many times in algebraic fraction problems, after multiplying by the LCD and simplifying, you'll find yourself with a quadratic equation. By isolating the variable term \(x\), we can determine the possible solutions that satisfy the initial equation, as seen in the exercise provided.
Algebraic Manipulation
Algebraic manipulation involves the use of various arithmetic operations and properties to simplify expressions or solve equations. Key skills include expanding polynomials, factoring expressions, combining like terms, and moving terms from one side of an equation to the other to isolate the variable in question.

In our exercise example, after multiplying through by the LCD, we use these algebraic manipulation techniques to simplify the equation and combine like terms to arrive at a solvable format. Mastery of algebraic manipulation is crucial to progressing in mathematics, as it lays the foundation for resolving complex problems.
Extraneous Solutions
An extraneous solution is a 'false' solution that emerges from the algebraic process of solving an equation but does not satisfy the original equation. They often arise when we multiply or square both sides of an equation, creating potential solutions that aren't actual solutions to the equation we started with. In the context of algebraic fractions, checking for extraneous solutions involves substituting the obtained values back into the original equation and ensuring that they don't make any denominators equal zero (which would invalidate the solution).

In the exercise, we validate that the solution \(x = 17\) is not extraneous by confirming that it does not cause division by zero in the original fractions. This verification step is a critical part of the solution process to ensure that answers are meaningful within their mathematical context.

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