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

STEP 1

1. We are dealing with algebraic equations involving fractions.
2. We will need to isolate the variable $x$ in each equation.
3. We will check each solution to ensure it satisfies the original equation.

STEP 2

1. Solve equation (a) for $x$ .
2. Solve equation (b) for $x$ .
3. Solve equation (c) for $x$ .
4. Check each solution to verify correctness.

STEP 3

For equation (a) $\frac{x-3}{10} = 4x$ , first eliminate the fraction by multiplying both sides by 10:
$\frac{x-3}{10} = 4x$ $x - 3 = 40x$

STEP 4

Rearrange the equation to isolate $x$ :
$x - 40x = 3$ $-39x = 3$
Divide both sides by $-39$ :
$x = -\frac{3}{39}$ $x = -\frac{1}{13}$

STEP 5

For equation (b) $\frac{3}{x} - 2 = \frac{5}{x}$ , first get a common denominator:
$\frac{3}{x} - \frac{5}{x} = 2$ $\frac{3 - 5}{x} = 2$ $\frac{-2}{x} = 2$

STEP 6

Cross-multiply to solve for $x$ :
$-2 = 2x$
Divide both sides by 2:
$x = -1$

STEP 7

For equation (c) $\frac{3}{x+2} - \frac{1}{x} = \frac{1}{5x}$ , first find a common denominator for the left side:
$\frac{3x - (x+2)}{x(x+2)} = \frac{1}{5x}$ $\frac{3x - x - 2}{x(x+2)} = \frac{1}{5x}$ $\frac{2x - 2}{x(x+2)} = \frac{1}{5x}$

STEP 8

Cross-multiply to solve for $x$ :
$5x(2x - 2) = x(x + 2)$ $10x^2 - 10x = x^2 + 2x$
Rearrange the equation:
$10x^2 - 10x - x^2 - 2x = 0$ $9x^2 - 12x = 0$
Factor out the common term:
$3x(3x - 4) = 0$
Set each factor to zero:
$3x = 0 \quad \text{or} \quad 3x - 4 = 0$ $x = 0 \quad \text{or} \quad x = \frac{4}{3}$

STEP 9

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) $x = \frac{4}{3}$

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