Answer:
[tex]\dfrac{7}{8} \textsf{ and } -\dfrac{5}{8}[/tex]
Step-by-step explanation:
To find numbers that are [tex]\dfrac{3}{4}[/tex] units distance from [tex]\displaystyle \dfrac{1}{8}[/tex] , add and subtract [tex]\dfrac{3}{4}[/tex] to and from [tex]\displaystyle \dfrac{1}{8}[/tex]
One of the points is at [tex]\displaystyle \dfrac{1}{8} + \dfrac{3}{4}[/tex] units away and the other at [tex]\displaystyle \dfrac{1}{8} - \dfrac{3}{4}[/tex] units away
- Compute [tex]\displaystyle \dfrac{1}{8} + \dfrac{3}{4} :[/tex]
[tex]\mathrm{For}\:\frac{3}{4}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:2[/tex]:
[tex]\frac{3}{4}=\frac{3\cdot \:2}{4\cdot \:2}=\frac{6}{8}[/tex]
[tex]\displaystyle \dfrac{1}{8} + \dfrac{3}{4} = \frac{1}{8}+\frac{6}{8} = \frac{7}{8}[/tex]
- Compute [tex]\frac{1}{8}-\frac{3}{4}[/tex]
Again for [tex]\frac{3}{4}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:2[/tex]
[tex]\frac{3}{4}=\frac{3\cdot \:2}{4\cdot \:2}=\frac{6}{8}[/tex]
[tex]\frac{1}{8}-\frac{3}{4} = \frac{1-6}{8} = \frac{-5}{8}[/tex]
