X.

Multiply by the L. C. D., $x^2-1$, and we have,

Exercise 51.

Solve:

1. ${\displaystyle \frac{x-1}{2}}={\displaystyle \frac{x+1}{3}}$.

2. ${\displaystyle \frac{3x-1}{4}}={\displaystyle \frac{2x+1}{3}}$.

3. ${\displaystyle \frac{6x-19}{2}}={\displaystyle \frac{2x-11}{3}}$.

4. ${\displaystyle \frac{7x-40}{8}}={\displaystyle \frac{9x-80}{10}}$.

5. ${\displaystyle \frac{3x-116}{4}}+{\displaystyle \frac{180-5x}{6}}=0$.

6. ${\displaystyle \frac{3x-4}{2}}-{\displaystyle \frac{3x-1}{16}}={\displaystyle \frac{6x-5}{8}}$.

7. ${\displaystyle \frac{x-1}{8}}-{\displaystyle \frac{x+1}{18}}=1$.

8. ${\displaystyle \frac{60-x}{14}}-{\displaystyle \frac{3x-5}{7}}={\displaystyle \frac{3x}{4}}$.

9. ${\displaystyle \frac{3x-1}{11}}-{\displaystyle \frac{2-x}{10}}={\displaystyle \frac{6}{5}}$.

10. ${\displaystyle \frac{4x}{x+1}}-{\displaystyle \frac{x}{x-2}}=3$.

11. ${\displaystyle \frac{2x+1}{4}}-{\displaystyle \frac{4x-1}{10}}+1-\frac{1}{4}=0$.

12. ${\displaystyle \frac{x-1}{5}}-{\displaystyle \frac{43-5x}{6}}-{\displaystyle \frac{3x-1}{8}}=0$.

13. ${\displaystyle \frac{1}{x+7}}={\displaystyle \frac{2}{x+1}}-{\displaystyle \frac{1}{x+3}}$.

14. ${\displaystyle \frac{1}{x+4}}+{\displaystyle \frac{2}{x+6}}-{\displaystyle \frac{3}{x+5}}=0$.

15. ${\displaystyle \frac{4}{x^2-1}}+{\displaystyle \frac{1}{x-1}}+{\displaystyle \frac{1}{x+1}}=0$.

16. ${\displaystyle \frac{3x+1}{4}}-{\displaystyle \frac{5x-4}{7}}=12-2x-{\displaystyle \frac{x-2}{3}}$.

17. $\frac{1}{8}(5x+3)-\frac{1}{3}(3-4x)+\frac{1}{6}(9-5x)=\frac{1}{2}(31-x)$.

18. $\frac{1}{15}(34x-56)-\frac{1}{5}(7x-3)-\frac{1}{3}(7x-5)=0$.