Solve Linear Equation 12Ecbc

1. **State the problem:** Solve the equation $$-\frac{7}{3} - \frac{1}{4} u = \frac{1}{4}$$ for $u$. 2. **Isolate the term with $u$:** Add $$\frac{7}{3}$$ to both sides to move the constant term. $$-\frac{7}{3} - \frac{1}{4} u + \frac{7}{3} = \frac{1}{4} + \frac{7}{3}$$ Simplifies to: $$-\frac{1}{4} u = \frac{1}{4} + \frac{7}{3}$$ 3. **Find common denominator and add the fractions on the right:** The common denominator of 4 and 3 is 12. $$\frac{1}{4} = \frac{3}{12}, \quad \frac{7}{3} = \frac{28}{12}$$ So, $$\frac{3}{12} + \frac{28}{12} = \frac{31}{12}$$ Thus, $$-\frac{1}{4} u = \frac{31}{12}$$ 4. **Solve for $u$ by dividing both sides by $$-\frac{1}{4}$$:** Dividing by a fraction is the same as multiplying by its reciprocal. $$u = \frac{31}{12} \div \left(-\frac{1}{4}\right) = \frac{31}{12} \times \left(-4\right)$$ 5. **Simplify the multiplication:** $$u = \frac{31 \times (-4)}{12} = \frac{-124}{12}$$ 6. **Simplify the fraction by dividing numerator and denominator by their greatest common divisor (4):** $$u = \frac{\cancel{4} \times (-31)}{\cancel{4} \times 3} = -\frac{31}{3}$$ **Final answer:** $$u = -\frac{31}{3}$$