Questions: Solve the inequality (u+frac110<frac320), and write the solution in interval notation.

Solve the inequality (u+frac110<frac320), and write the solution in interval notation.
Transcript text: Solve the inequality $u+\frac{1}{10}<\frac{3}{20}$, and write the solution in interval notation.
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Solution

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Solution Steps

To solve the inequality \( u + \frac{1}{10} < \frac{3}{20} \), we need to isolate the variable \( u \). This involves subtracting \(\frac{1}{10}\) from both sides of the inequality. After simplifying the right side, we can express the solution in interval notation.

Step 1: Isolate the Variable

We start with the inequality:

\[ u + \frac{1}{10} < \frac{3}{20} \]

To isolate \( u \), we need to subtract \(\frac{1}{10}\) from both sides of the inequality:

\[ u < \frac{3}{20} - \frac{1}{10} \]

Step 2: Find a Common Denominator

The fractions \(\frac{3}{20}\) and \(\frac{1}{10}\) need a common denominator to be subtracted. The least common denominator of 20 and 10 is 20. Convert \(\frac{1}{10}\) to a fraction with a denominator of 20:

\[ \frac{1}{10} = \frac{2}{20} \]

Step 3: Subtract the Fractions

Now subtract the fractions:

\[ u < \frac{3}{20} - \frac{2}{20} = \frac{1}{20} \]

Final Answer

The solution to the inequality is:

\[ u < \frac{1}{20} \]

In interval notation, this is expressed as:

\[ \boxed{(-\infty, \frac{1}{20})} \]

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