Question2. Solve algebraically. Check each solution. a) $\frac{x-3}{10}=4 x$ b) $\frac{3}{x}-2=\frac{5}{x}$ c) $\frac{3}{x+2}-\frac{1}{x}=\frac{1}{5 x}$

Studdy Solution
Check each solution by substituting back into the original equations:
For (a), substitute $x = -\frac{1}{13}$ into $\frac{x-3}{10} = 4x$: $$\frac{-\frac{1}{13} - 3}{10} = 4(-\frac{1}{13})$$ $$\frac{-\frac{40}{13}}{10} = -\frac{4}{13}$$ $$-\frac{4}{13} = -\frac{4}{13}$$ (True)
For (b), substitute $x = -1$ into $\frac{3}{x} - 2 = \frac{5}{x}$: $$\frac{3}{-1} - 2 = \frac{5}{-1}$$ $$-3 - 2 = -5$$ $$-5 = -5$$ (True)
For (c), substitute $x = 0$ and $x = \frac{4}{3}$ into $\frac{3}{x+2} - \frac{1}{x} = \frac{1}{5x}$: $x = 0$ is not valid as it causes division by zero. Substitute $x = \frac{4}{3}$: $$\frac{3}{\frac{4}{3} + 2} - \frac{1}{\frac{4}{3}} = \frac{1}{5(\frac{4}{3})}$$ $$\frac{3}{\frac{10}{3}} - \frac{3}{4} = \frac{3}{20}$$ $$\frac{9}{10} - \frac{3}{4} = \frac{3}{20}$$ $$\frac{18}{20} - \frac{15}{20} = \frac{3}{20}$$ $$\frac{3}{20} = \frac{3}{20}$$ (True)
The solutions are: a) $x = -\frac{1}{13}$ b) $x = -1$ c) $x = \frac{4}{3}$