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Chapter 7: Problem 2

Solve each rational equation. $$\frac{x}{5}=\frac{x}{6}+1$$

Short Answer

Expert verified
The solution to the rational equation is \( x = 30 \)

Step by step solution

01

Eliminate the denominator

We will start to eliminate the denominator by multiplying every term in the equation by \(30\) (Least common multiple of \(5\) and \(6\)). This gives us: \(6x = 5x + 30\)
02

Simplify the equation

Next, let's simplify this equation by subtracting \(5x\) from both sides of the equation. This gives: \(x = 30\)
03

Verify the solution

Finally, we verify our solution by substituting \(x = 30\) into the original equation and confirming both sides of the equation are equal. Substituting \(x = 30\) we get: \(\frac{30}{5}=\frac{30}{6}+ 1\), simplifying the left-side gives us \(6 = 5 + 1\), confirming our solution is correct since both sides of the equation equal to \(6\).

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Key Concepts

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

Least Common Multiple
When dealing with rational equations, like \(\frac{x}{5}=\frac{x}{6}+1\), it's often useful to eliminate denominators to simplify calculations. The key to doing this effectively is by using the Least Common Multiple (LCM). The LCM of two numbers is the smallest number that is evenly divisible by both. In our case, the denominators are 5 and 6.
  • To find the LCM of 5 and 6, list the multiples of each number:
  • Multiples of 5: 5, 10, 15, 20, 25, 30, ...
  • Multiples of 6: 6, 12, 18, 24, 30, ...
The smallest common multiple is 30, which is why we choose to multiply every term in the equation by 30. This action effectively removes the denominators, turning the equation into an easier form, making it simpler to solve.
Solving Equations
Once you've cleared the denominators, you can proceed to simplify the rational equation into a linear equation. For our equation \(6x = 5x + 30\), you need to solve for \(x\). These steps often involve basic operations like addition, subtraction, multiplication, and division.
  • Begin by isolating the variable \(x\). In this instance, subtract 5x from both sides, which gives \(x = 30\).
  • The goal is to have the variable on one side of the equation and numbers on the other.
It's crucial to simplify each step systematically to prevent mistakes and ensure accuracy.
Verifying Solutions
After solving the equation, it's essential to verify that your solution is correct. This verification step ensures the solution satisfies the original equation. To do this, substitute \(x = 30\) back into the original rational equation \(\frac{x}{5}=\frac{x}{6}+1\).
  • On the left side of the equation, substituting gives \(\frac{30}{5} = 6\).
  • On the right side, substituting gives \(\frac{30}{6} + 1 = 5 + 1 = 6\).
Since both sides of the equation are equal at \(6\), this confirms that \(x = 30\) is indeed the correct solution. Verifying your solution is crucial as it reinforces the correctness of your work, providing assurance and confidence in your problem-solving process.

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Most popular questions from this chapter

What are similar triangles?

Solve: \(x^{2}-12 x+36=0 .\) (Section 6.6, Example 4)

Describe two similarities between the following problems: $$ \frac{3}{8}+\frac{1}{8} \text { and } \frac{x}{x^{2}-1}+\frac{1}{x^{2}-1} $$

perform the indicated operations. Simplify the result, if possible. $$\left(\frac{3 x-1}{x^{2}+5 x-6}-\frac{2 x-7}{x^{2}+5 x-6}\right) \div \frac{x+2}{x^{2}-1}$$

perform the indicated operation or operations. Simplify the result, if possible. $$\frac{6 b^{2}-10 b}{16 b^{2}-48 b+27}+\frac{7 b^{2}-20 b}{16 b^{2}-48 b+27}-\frac{6 b-3 b^{2}}{16 b^{2}-48 b+27}$$
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