Given:
[tex]\frac{1}{4}|+\frac{1}{8}|+n+\frac{1}{2}|=1[/tex]
To Determine: The value of n
Solution:
[tex]\begin{gathered} \frac{1}{4}|+\frac{1}{8}|+\frac{1}{2}|+n=1 \\ \frac{2+1+4}{8}|+n=1 \\ \frac{7}{8}|+n=1 \end{gathered}[/tex][tex]\begin{gathered} n=1-\frac{7}{8}| \\ n=\frac{8-7}{8}| \\ n=\frac{1}{8}| \end{gathered}[/tex]
Hence, the value of n is
[tex]n=\frac{1}{8}|[/tex]
