Solve for b: \(\frac{\mathrm{1} }{\mathrm{2}}b+2= \frac{\mathrm{3} }{\mathrm{4}} b\)

Given \(\frac{\mathrm{1} }{\mathrm{2}}b+2= \frac{\mathrm{3} }{\mathrm{4}} b\)
Subtracting \(\frac{\mathrm{1} }{\mathrm{2}}b\) from both sides we get \(\frac{\mathrm{1}}{\mathrm{2}}b+2-\frac{\mathrm{1}}{\mathrm{2}}b=\frac{\mathrm{3}}{\mathrm{4}}b-\frac{\mathrm{1}}{\mathrm{2}}b \Rightarrow 2=\frac{\mathrm{3} }{\mathrm{4}}b- \frac{\mathrm{1x2} }{\mathrm{2x2}}b\) (changing \(\frac{\mathrm{3} }{\mathrm{4}}\) and \(\frac{\mathrm{1} }{\mathrm{2}}\) to the same denominators)
\(2= \frac{\mathrm{3} }{\mathrm{4}}b-\frac{\mathrm{2} }{\mathrm{4}}b=2 \Rightarrow b=4 \times 2=8\)
\(=>b=8\)