Answer :
[tex]\frac{1}{2}r+2(\frac{3}{4}r-1)=\frac{1}{4}r+6\ \ \ \ \ \ \ |omit\ bracket\\\\\frac{1}{2}r+\frac{3}{2}r-2=\frac{1}{4}r+6\ \ \ \ \ \ |\cdot4\\\\2r+6r-8=r+24\ \ \ \ \ \ |combine\ like\ terms\\\\2r+6r-r=24+8\ \ \ \ \ |simplify\\\\7r=32\ \ \ \ \ |:7\\\\r=\frac{32}{7}=4\frac{4}{7}[/tex]
[tex]\frac{1}{2}r+2(\frac{3}{4}r-1)=\frac{1}{4}r+6\\\\\frac{1}{2}r+2\times\frac{3}{4}r-2\cdot1=\frac{1}{4}r+6\\\\\frac{1}{2}r+\frac{3}{2}r-2=\frac{1}{4}r+6\\\\\frac{1+3}{2}r-2=\frac{1}{4}r+6\\\\\frac{4}{2}r-2=\frac{1}{4}r+6\\\\2r-2=\frac{1}{4}r+6\ \ \ \ \ \ |add\ 2\ to\ both\ sides\\\\2r=\frac{1}{4}r+8\ \ \ \ |subtract\ \frac{1}{4}r\ from\ both\ sides[/tex]
[tex]1\frac{3}{4}r=8\\\\\frac{1\cdot4+3}{4}r=8\\\\\frac{7}{4}r=8\ \ \ \ \ |divide\ both\ sides\ by\ \frac{7}{4}\ (multiply\ both\ sides\ by\ \frac{4}{7})\\\\r=8\cdot\frac{4}{7}\\\\r=\frac{8\cdot4}{7}\\\\r=\frac{32}{7}\\\\\boxed{r=4\frac{4}{7}}[/tex]
[tex]1\frac{3}{4}r=8\\\\\frac{1\cdot4+3}{4}r=8\\\\\frac{7}{4}r=8\ \ \ \ \ |divide\ both\ sides\ by\ \frac{7}{4}\ (multiply\ both\ sides\ by\ \frac{4}{7})\\\\r=8\cdot\frac{4}{7}\\\\r=\frac{8\cdot4}{7}\\\\r=\frac{32}{7}\\\\\boxed{r=4\frac{4}{7}}[/tex]
