Chapter 11: Problem 34
Equations with Unknown in Denominator. \(\frac{x}{3}-\frac{x^{2}-5 x}{3 x-7}=\frac{2}{3}\)
Short Answer
Expert verified
The solution to the equation is \(x = -7\), provided we check that this value does not result in a zero denominator in the original equation.
Step by step solution
01
Find a common denominator
To combine the fractions on the left side of the equation, find a common denominator. Here, the common denominator is the product of the distinct denominatorial terms, which is \(3(3x-7)\).
02
Rewrite the equation with the common denominator
Express each fraction with the common denominator \(3(3x-7)\). This gives us \(\frac{x(3x-7)}{3(3x-7)} - \frac{(x^2-5x)(3)}{3(3x-7)} = \frac{2}{3}\). Simplify by distributing the numerators.
03
Simplify the equation
Combine the terms in the numerator on the left side of the equation, which results in \(\frac{3x^2 - 7x - 3x^2 + 15x}{3(3x-7)} = \frac{2}{3}\). Further simplification of the numerator gives \(\frac{8x}{3(3x-7)}\).
04
Remove the fraction by multiplying both sides by the denominator
To remove the fraction, multiply both sides of the equation by the common denominator \(3(3x-7)\). We get \(8x = 2(3x-7)\).
05
Distribute and simplify the equation
Distribute the right side of the equation to get \(8x = 6x - 14\). This simplifies to \(2x = -14\) after subtracting \(6x\) from both sides.
06
Solve for x
Divide both sides of \(2x = -14\) by 2 to isolate x. The solution is \(x = -7\).
07
Check solution in the original equation
Substitute \(x = -7\) into the original equation and simplify to verify the solution. Note: \(x\) cannot be \(7/3\), as this would make the denominator of the second term of the original equation equal to zero, which is undefined.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Common Denominators
When we work with equations involving fractions, the term 'common denominators' refers to a shared denominator that all fractions in the equation can use. It is a foundational step in solving fractional equations because it allows us to combine the fractions.
In our exercise, we have fractions with different denominators. To solve the equation, we need to be able to combine them, and for that, each term must have the same denominator. We identify the distinct denominators, in this case, 3 and 3x-7to find that the common denominator is their product, which is (3cdot(3x-7)). Rewriting each fraction with the common denominator allows us to work with a single term instead of multiple fractional terms, simplifying our equation significantly.
In our exercise, we have fractions with different denominators. To solve the equation, we need to be able to combine them, and for that, each term must have the same denominator. We identify the distinct denominators, in this case, 3 and 3x-7to find that the common denominator is their product, which is (3cdot(3x-7)). Rewriting each fraction with the common denominator allows us to work with a single term instead of multiple fractional terms, simplifying our equation significantly.
Fractional Equations
Fractional equations are equations that include one or more fractions. The variables can be found either in the numerators or denominators. Solving these equations typically involves finding a common denominator, which helps to eliminate the fractions and streamline the equation, as we have done in the given exercise.
To solve fractional equations efficiently, we combine all terms over the common denominator and then multiply through by that common denominator to clear the fractions. This leaves us with a simpler, often non-fractional, equation to solve—a crucial technique when facing these equations.
To solve fractional equations efficiently, we combine all terms over the common denominator and then multiply through by that common denominator to clear the fractions. This leaves us with a simpler, often non-fractional, equation to solve—a crucial technique when facing these equations.
Simplifying Algebraic Expressions
In algebra, simplifying expressions is a key skill. It involves reducing an expression to its simplest form, by performing operations like distributing multiplication over addition and subtraction, combining like terms, and eliminating unnecessary terms.
The equation from our example required distributing terms in the numerator, where we multiplied x by (3x-7)and (x^2-5x)by 3. After consolidation, we combined like terms which, in this case, were the terms involving x's squared, to simplify the numerator. This series of steps transformed a complex fractional equation into a more manageable linear equation.
The equation from our example required distributing terms in the numerator, where we multiplied x by (3x-7)and (x^2-5x)by 3. After consolidation, we combined like terms which, in this case, were the terms involving x's squared, to simplify the numerator. This series of steps transformed a complex fractional equation into a more manageable linear equation.
Verifying Solutions
After solving an equation, it's crucial to verify that our solutions are correct and that they do not create any undefined conditions, such as a zero in the denominator. Verifying involves substituting the found solution back into the original equation to ensure it holds true.
In our exercise, after finding the solution (x = -7), it needs to be checked in the original equation. This step confirms that the solution makes each side of the equation equal when simplified. Also, we need to ensure that the solution doesn't lead to any divisions by zero in the original equation, which would invalidate it. For instance, if x were 7/3, the denominator of the second term in the original equation would be zero, which cannot occur. Therefore, the solution must be disqualified from the set of possible solutions.
In our exercise, after finding the solution (x = -7), it needs to be checked in the original equation. This step confirms that the solution makes each side of the equation equal when simplified. Also, we need to ensure that the solution doesn't lead to any divisions by zero in the original equation, which would invalidate it. For instance, if x were 7/3, the denominator of the second term in the original equation would be zero, which cannot occur. Therefore, the solution must be disqualified from the set of possible solutions.