Question

2.08 - solve linear equations with fractions
due sunday by 11:59pm points 100 submitting an external tool available sep 8
question
solve: $-\frac{4}{7}b + \frac{1}{4}=\frac{1}{3}b + \frac{1}{2}$.
provide your answer below:
b =

Explanation:

Step1: Move terms with b to one - side

Subtract $\frac{1}{3}b$ from both sides and subtract $\frac{1}{4}$ from both sides:
$-\frac{4}{7}b-\frac{1}{3}b=\frac{1}{2}-\frac{1}{4}$

Step2: Find a common denominator for b - terms and constant terms

The common denominator of 7 and 3 is 21, and the common denominator of 2 and 4 is 4.
$(-\frac{4\times3}{7\times3}-\frac{1\times7}{3\times7})b=\frac{1\times2}{2\times2}-\frac{1}{4}$
$(-\frac{12}{21}-\frac{7}{21})b=\frac{2}{4}-\frac{1}{4}$

Step3: Simplify the fractions

$-\frac{12 + 7}{21}b=\frac{2-1}{4}$
$-\frac{19}{21}b=\frac{1}{4}$

Step4: Solve for b

Multiply both sides by $-\frac{21}{19}$:
$b=\frac{1}{4}\times(-\frac{21}{19})$
$b =-\frac{21}{76}$

Answer:

$b =-\frac{21}{76}$