Subjects algebra

Fraction Expression A39D3A

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1. **State the problem:** Simplify the expression $$\left(-\frac{3}{7} + \frac{1}{3}\right) - \left[ 2 - \left(3 - \frac{1}{7}\right) + \left(1 - \frac{7}{3}\right) \right].$$ 2. **Simplify inside the first parentheses:** $$-\frac{3}{7} + \frac{1}{3} = \frac{-3 \times 3}{7 \times 3} + \frac{1 \times 7}{3 \times 7} = \frac{-9}{21} + \frac{7}{21} = \frac{-9 + 7}{21} = \frac{-2}{21}.$$ 3. **Simplify inside the brackets step-by-step:** - Simplify $$3 - \frac{1}{7} = \frac{3 \times 7}{7} - \frac{1}{7} = \frac{21}{7} - \frac{1}{7} = \frac{20}{7}.$$ - Simplify $$1 - \frac{7}{3} = \frac{1 \times 3}{3} - \frac{7}{3} = \frac{3}{3} - \frac{7}{3} = \frac{3 - 7}{3} = \frac{-4}{3}.$$ 4. **Substitute back into the bracket:** $$2 - \left(3 - \frac{1}{7}\right) + \left(1 - \frac{7}{3}\right) = 2 - \frac{20}{7} + \left(-\frac{4}{3}\right).$$ 5. **Convert all terms to a common denominator to combine:** - Convert 2 to $$\frac{42}{21}$$ (since 21 is the LCM of 7 and 3). - Convert $$\frac{20}{7}$$ to $$\frac{60}{21}$$. - Convert $$-\frac{4}{3}$$ to $$-\frac{28}{21}$$. 6. **Combine:** $$\frac{42}{21} - \frac{60}{21} - \frac{28}{21} = \frac{42 - 60 - 28}{21} = \frac{-46}{21}.$$ 7. **Now substitute back into the original expression:** $$\left(-\frac{2}{21}\right) - \left(-\frac{46}{21}\right) = -\frac{2}{21} + \frac{46}{21} = \frac{-2 + 46}{21} = \frac{44}{21}.$$ **Final answer:** $$\boxed{\frac{44}{21}}.$$