Hello, mndn6154!
A few
spaces would make it easier to read . . .
I can only
guess at what you meant to say . . .
We have: \(\displaystyle \:\frac{1}{2}|6\,-\,x|\,-\,\frac{3}{4}\;>\;1\) . . .
I think!
Add \(\displaystyle \frac{3}{4}\) to both sides: \(\displaystyle \;\frac{1}{2}|6\,-\,x|\;>\;\frac{7}{4}\)
Multiply by 2: \(\displaystyle \;|6\,-\,x|\;>\;\frac{7}{2}\)
Then: \(\displaystyle \;-\frac{7}{2}\;>\; 6\,-\,x\;>\;\frac{7}{2}\)
Subtract \(\displaystyle 6:\;\;-\frac{19}{2}\;>\;-x\;>\;-\frac{5}{2}\)
Multiply by -1: \(\displaystyle \:\frac{19}{2}\;<\:x\;<\;\frac{5}{2}\)
Therefore: \(\displaystyle \,x\,<\,\frac{5}{2}\,\)
or \(\displaystyle \,x\, >\, \frac{19}{2}\)
Interval notation: \(\displaystyle \,\left(-\infty,\,\frac{5}{2}\right)\:\cup\:\left(\frac{19}{2}\,\infty\right)\)
