Question

solve for b.\\(\frac{1}{3} - \frac{3}{2}b + \frac{3}{4} = -\frac{5}{6}b\\)\\(b = \\)

Explanation:

Step 1: Combine constant terms

First, add the constant terms \(\frac{1}{3}\) and \(\frac{3}{4}\). 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 \(\frac{13}{12}-\frac{3}{2}b = -\frac{5}{6}b\).

Step 2: Get all b terms on one side

Add \(\frac{3}{2}b\) to both sides to get \(\frac{13}{12}=-\frac{5}{6}b+\frac{3}{2}b\). 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 \(\frac{13}{12}=\frac{2}{3}b\).

Step 3: 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: \(\frac{13\times3}{12\times2}=\frac{39}{24}=\frac{13}{8}\). Wait, no, wait: Wait, \(\frac{13}{12}\times\frac{3}{2}\): the 3 and 12 cancel: 12÷3=4, so \(\frac{13}{4}\times\frac{1}{2}=\frac{13}{8}\)? Wait, no, 13×3=39, 12×2=24, 39÷3=13, 24÷3=8. Yes, so \(b=\frac{13}{8}\)? Wait, no, let's check again. Wait, step 2: \(-\frac{5}{6}b+\frac{3}{2}b\): \(\frac{3}{2}b\) is \(\frac{9}{6}b\), so \(-\frac{5}{6}+\frac{9}{6}=\frac{4}{6}=\frac{2}{3}\), correct. Then step 3: \(\frac{13}{12}=\frac{2}{3}b\), so \(b=\frac{13}{12}\div\frac{2}{3}=\frac{13}{12}\times\frac{3}{2}=\frac{13\times3}{12\times2}=\frac{39}{24}=\frac{13}{8}\). Wait, but let's check the original equation. Let's plug \(b = \frac{13}{8}\) back in. Left side: \(\frac{1}{3}-\frac{3}{2}\times\frac{13}{8}+\frac{3}{4}\). \(\frac{3}{2}\times\frac{13}{8}=\frac{39}{16}\). \(\frac{1}{3}=\frac{16}{48}\), \(\frac{39}{16}=\frac{117}{48}\), \(\frac{3}{4}=\frac{36}{48}\). So \(\frac{16}{48}-\frac{117}{48}+\frac{36}{48}=\frac{16 - 117 + 36}{48}=\frac{-65}{48}\). Right side: \(-\frac{5}{6}\times\frac{13}{8}=-\frac{65}{48}\). Oh, so that's correct. Wait, but earlier when I did step 1, I think I made a mistake in the constant term. Wait, \(\frac{1}{3}+\frac{3}{4}\): common denominator 12, so \(\frac{4}{12}+\frac{9}{12}=\frac{13}{12}\), correct. Then equation is \(\frac{13}{12}-\frac{3}{2}b = -\frac{5}{6}b\). Then add \(\frac{3}{2}b\) to both sides: \(\frac{13}{12}=-\frac{5}{6}b+\frac{3}{2}b\). Convert \(\frac{3}{2}b\) to sixths: \(\frac{9}{6}b\), so \(-\frac{5}{6}+\frac{9}{6}=\frac{4}{6}=\frac{2}{3}\), so \(\frac{13}{12}=\frac{2}{3}b\). Then multiply both sides by \(\frac{3}{2}\): \(b=\frac{13}{12}\times\frac{3}{2}=\frac{13\times3}{12\times2}=\frac{39}{24}=\frac{13}{8}\). Yes, that's correct. Wait, but let me check again. Wait, \(\frac{13}{12}\times\frac{3}{2}\): 3 and 12 cancel, 12÷3=4, so \(\frac{13}{4}\times\frac{1}{2}=\frac{13}{8}\). Correct. So the value of b is \(\frac{13}{8}\)? Wait, no, wait, when I plugged back in, left side: \(\frac{1}{3}-\frac{3}{2}\times\frac{13}{8}+\frac{3}{4}\). Let's compute each term: \(\frac{1}{3}=\frac{8}{24}\), \(\frac{3}{2}\times\frac{13}{8}=\frac{39}{16}=\frac{58.5}{24}\)? Wait, no, better to use sixteenths? Wait, no, let's use common denominator 48. \(\frac{1}{3}=\frac{16}{48}\), \(\frac{3}{2}b=\frac{3}{2}\times\frac{13}{8}=\frac{39}{16}=\frac{117}{48}\), \(\frac{3}{4}=\frac{36}{48}\). So \(\frac{16}{48}-\frac{117}{48}+\frac{36}{48}=\frac{16 + 36 - 117}{48}=\frac{52 - 117}{48}=\frac{-65}{48}\). Right side: \(-\frac{5}{6}\times\frac{13}{8}=-\frac{65}{48}\). Yes, that's correct. So b is \(\frac{13}{8}\). Wait, but let me check the arithmetic again. Step 1: \(\frac{1}{3}+\frac{3}{4}\). 3 and 4 have LCM 12. \(\frac{1\times4}{3\times4}=\frac{4}{12}\), \(\frac{3\times3}{4\times3}=\frac{9}{12}\). 4+9=13, so \(\frac{13}{12}\). Correct. Step 2: \(-\frac{3}{2}b + \frac{5}{6}b\)? Wait, no, original equation: \(\frac{13}{12}-\frac{3}{2}b = -\frac{5}{6}b\). So to get rid of the negative on the right, add \(\frac{3}{2}b\) to both sides: \(\frac{13}{12}=-\frac{5}{6}b+\frac{3}{2}b\). Then \(\frac{3}{2}b\) is \(\frac{9}{6}b\), so \(-\frac{5}{6}b+\frac{9}{6}b=\frac{4}{6}b=\frac{2}{3}b\). Correct. Then \(\frac{13}{12}=\frac{2}{3}b\). Multiply both sides by \(\frac{3}{2}\): \(b=\frac{13}{12}\times\frac{3}{2}=\frac{13\times3}{12\times2}=\frac{39}{24}=\frac{13}{8}\). Yes, that's correct. So the answer is \(\frac{13}{8}\). Wait, but let me check once more. Maybe I made a mistake in the sign. Original equation: \(\frac{1}{3}-\frac{3}{2}b+\frac{3}{4}=-\frac{5}{6}b\). Combine constants: \(\frac{1}{3}+\frac{3}{4}=\frac{4 + 9}{12}=\frac{13}{12}\). So \(\frac{13}{12}-\frac{3}{2}b = -\frac{5}{6}b\). Then, add \(\frac{3}{2}b\) to both sides: \(\frac{13}{12}=-\frac{5}{6}b+\frac{3}{2}b\). Now, \(\frac{3}{2}b\) is \(\frac{9}{6}b\), so \(-\frac{5}{6}b+\frac{9}{6}b=\frac{4}{6}b=\frac{2}{3}b\). So \(\frac{13}{12}=\frac{2}{3}b\). Then, multiply both sides by \(\frac{3}{2}\): \(b=\frac{13}{12}\times\frac{3}{2}=\frac{13\times3}{12\times2}=\frac{39}{24}=\frac{13}{8}\). Yes, that's correct. So the solution is \(b = \frac{13}{8}\). Wait, but let me check with another approach. Multiply all terms by 12 to eliminate denominators. 12×\(\frac{1}{3}\)=4, 12×(-\(\frac{3}{2}\)b)=-18b, 12×\(\frac{3}{4}\)=9, 12×(-\(\frac{5}{6}\)b)=-10b. So the equation becomes 4 - 18b + 9 = -10b. Combine constants: 4 + 9 = 13. So 13 - 18b = -10b. Add 18b to both sides: 13 = 8b. Then b = \(\frac{13}{8}\). Oh! Wait, that's a simpler way. Multiply all terms by 12 (LCM of 3, 2, 4, 6). So 12×\(\frac{1}{3}\)=4, 12×(-\(\frac{3}{2}\)b)=-18b, 12×\(\frac{3}{4}\)=9, 12×(-\(\frac{5}{6}\)b)=-10b. So 4 - 18b + 9 = -10b. 4 + 9 = 13, so 13 - 18b = -10b. Add 18b to both sides: 13 = 8b. Then b = \(\frac{13}{8}\). Ah, that's much simpler. I should have done that first. So that confirms it. So the correct answer is \(\frac{13}{8}\).

Answer:

\(\frac{13}{8}\)