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Question
solve for b.
\frac{1}{3} - \frac{3}{2}b + \frac{3}{4} = -\frac{5}{6}b
b =
Step1: Combine constant terms
First, add the constant terms \(\frac{1}{3}\) and \(\frac{3}{4}\). To add these fractions, find a common denominator, which is 12. So \(\frac{1}{3}=\frac{4}{12}\) and \(\frac{3}{4}=\frac{9}{12}\). Then \(\frac{4}{12}+\frac{9}{12}=\frac{13}{12}\). The equation becomes \(\frac{13}{12}-\frac{3}{2}b = -\frac{5}{6}b\).
Step2: Get all b terms on one side
Add \(\frac{3}{2}b\) to both sides of the equation. \(\frac{13}{12}=-\frac{5}{6}b+\frac{3}{2}b\). Now, convert \(\frac{3}{2}b\) to sixths: \(\frac{3}{2}b=\frac{9}{6}b\). So \(-\frac{5}{6}b+\frac{9}{6}b=\frac{4}{6}b=\frac{2}{3}b\). The equation is now \(\frac{13}{12}=\frac{2}{3}b\).
Step3: Solve for b
Multiply both sides by the reciprocal of \(\frac{2}{3}\), which is \(\frac{3}{2}\). So \(b = \frac{13}{12}\times\frac{3}{2}\). Simplify: \(\frac{13\times3}{12\times2}=\frac{39}{24}=\frac{13}{8}\).
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\(\frac{13}{8}\)