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Chapter 9: Problem 55

If the exercise is an equation, solve it and check. Otherwise, perform the indicated operations and simplify. $$\frac{x}{x-5}+\frac{3}{2}=\frac{5}{x-5}$$

Short Answer

Expert verified
There is no solution because \(x = 5\) leads to division by zero.

Step by step solution

01

Identify the common denominator

The common denominator for the fractions is \(x-5\). Rewrite each term with this common denominator.
02

Rewrite each fraction with a common denominator

\(\frac{x}{x-5} + \frac{3}{2} = \frac{5}{x-5}\) The middle term needs to be rewritten with the common denominator. Multiply \(\frac{3}{2}\) by \((x-5)/(x-5)\): \ \(\frac{x}{x-5} + \frac{3(x-5)}{2(x-5)} = \frac{5}{x-5}\)
03

Combine the fractions on the left-hand side

\(\frac{x \cdot 2 + 3(x-5)}{2(x-5)} = \frac{5}{x-5}\) Simplify the numerator: \ \(2x + 3x - 15 = 5x - 15\) Now the equation is \ \(\frac{5x - 15}{2(x-5)} = \frac{5}{x-5}\)
04

Clear the fractions by cross-multiplying

Cross-multiply to eliminate the denominators: \ \( (5x - 15)(x-5) = 10(x-5)\)
05

Simplify the equation

Simplify both sides of the equation: \ \(5x - 15 = 10\) Add 15 to both sides: \ \(5x = 25\) Divide by 5 on both sides: \ \(x = 5\)
06

Check the solution

Substitute \(x = 5\) back into the original equation to check: \( \frac{5}{0} + \frac{3}{2} = \frac{5}{0}\) This causes a division by zero, so \(x = 5\) is not a valid solution.
07

Conclude there is no solution

No other solutions are possible given \(x = 5\) results in an undefined value. Therefore, the equation has no solution.

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Key Concepts

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

common denominator
When solving rational equations, identifying a common denominator is crucial. It allows us to combine and compare fractions more easily. For the equation \(\frac{x}{x-5}+\frac{3}{2}=\frac{5}{x-5}\), we notice the denominators are \(x-5\) and \(2\). We need all fractions to share the same denominator. By multiplying the fraction \(\frac{3}{2}\) by \(\frac{x-5}{x-5}\), we rewrite it to also have \(x-5\) as the denominator. Now, all fractions share the common denominator \(x-5\), simplifying our next steps.
cross-multiplying
Cross-multiplying helps us eliminate fractions from both sides of the equation. After combining fractions on one side, our equation becomes \(\frac{5x - 15}{2(x-5)} = \frac{5}{x-5}\). To clear the fractions, we cross-multiply: \[ (5x - 15)(x-5) = 10(x-5) \]. This step ensures both sides of the equation are in a polynomial form, making them easier to handle. Make sure you properly distribute and simplify terms after cross-multiplying.
simplifying fractions
Simplifying fractions earlier in the problem can make equations more manageable. Once we鈥檝e rewritten fractions with a common denominator, we can combine them. For example, combining \(\frac{x}{x-5} + \frac{3(x-5)}{2(x-5)}\) gives \(\frac{ 2x + 3(x-5) }{ 2(x-5) } = \frac{5x - 15 }{ 2(x-5) }\). Simplifying \(2x + 3(x-5)\) to \(5x - 15\) reduces the problem's complexity. Make sure each step accurately reflects the algebraic operations performed.
checking solutions
Checking solutions is vital, especially for equations involving potential undefined values. For \(\frac{x}{x-5}+\frac{3}{2}=\frac{5}{x-5}\), solving eventually gave us \(x = 5\). By substituting back, we check: \(\frac{5}{0} + \frac{3}{2} = \frac{5}{0}\). This causes a division by zero, revealing \(x = 5\) is invalid. Always substitute your solutions back into the original equation to verify they're valid within the problem's constraints. No valid solution means the given equation has no solution.

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Most popular questions from this chapter

If the exercise is an equation, solve it and check. Otherwise, perform the indicated operations and simplify. $$\frac{5}{y^{2}-3 y-10}+\frac{3}{y-5}=\frac{-1}{4 y+8}$$

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