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Chapter 14: Problem 29

Solve and write the answer in set-builder notation. $$d+\frac{1}{2}<\frac{1}{3}$$

Short Answer

Expert verified
In set-builder notation, the solution is \(d : d < -\frac{1}{6}\).

Step by step solution

01

Isolate the term \(d\)

Subtract \(\frac{1}{2}\) from both sides of the inequality to isolate the term \(d\). This gives us the inequality: \(d<\frac{1}{3}-\frac{1}{2}\).
02

Simplify the right side

To facilitate further operations, find the common denominator and subtract the fractions on the right side. The common denominator of 2 and 3 is 6, so the inequality becomes: \(d<\frac{2}{6}-\frac{3}{6}\). Simplifying the fraction gives: \(d<-\frac{1}{6}\).
03

Write in set-builder notation

The solution to the inequality can be expressed in set-builder notation: \(d : d < -\frac{1}{6}\). This reads 'the set of all \(d\) such that \(d\) is less than -1/6.'

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