Solving Algebraic Fractions

Algebraic fractions involve variables in the numerator, denominator, or both. Solving equations with algebraic fractions requires careful handling of the fractions, often by eliminating them first. Let’s break it down step by step

 

Key Steps to Solve Algebraic Fractions

1. Eliminate the Denominators: Multiply through by the least common denominator (LCD) to remove the fractions.

2. Simplify the Equation: Expand and collect terms to form a simpler equation.

3. Solve for $x$: Use standard techniques like isolating $x$, factoring, or applying the quadratic formula.

4. Check for Restrictions: Ensure your solution does not make any denominator equal to zero (which would make the fraction undefined).

 

Example: Solving a Simple Fraction Equation

Solve: $$\frac{x}{4} + \frac{1}{2} = \frac{3}{4}$$

Solution:

1. Find the least common denominator (LCD): The LCD of $4$ and $2$ is $4$.

2. Multiply through by $4$: $$4 \cdot \frac{x}{4} + 4 \cdot \frac{1}{2} = 4 \cdot \frac{3}{4} \\ \text{Simplfy: }  \\  x + 2 = 3$$
3. Solve for $x$: $$x = 3 - 2 \\ x = 1 \\ \text{Final Answer: }  \\  x = 1$$ 

 

Example: Solving with Quadratic Expressions

Solve: $$\frac{x}{x - 2} = 3$$

Solution:

1. Multiply through by $x - 2$ (provided $x \not = 2$): $$(x - 2) \cdot \frac{x}{x - 2} = 3 \cdot (x - 2) \\ \text{Simplfy: } \\ x = 3(x - 2)$$

2. Expand and rearrange: $$x = 3x -6  \\  x - 3x = - 6 \\  -2x = -6$$
3. Solve for $x$: $$x = \frac{-6}{-2}  \\  x = 3$$
4. Check the restriction: $x \not = 2$, so $x = 3$ is valid

Final Answer: $$x = 3$$

 

 

 

 

Example 4: Solving with Multiple Denominators

Solve: $$\frac{2x}{x + 1} - \frac{1}{x - 1} = 1$$

Solution:

1. Find the LCD: The LCD of $x + 1$ and $x - 1$ is $(x + 1)(x - 1)$

2. Multiply through by $(x + 1)(x - 1)$: $$(x +1)(x - 1) \cdot \frac{2x}{x + 1} - (x + 1)(x - 1) \cdot \frac{1}{x - 1} = (x + 1)(x - 1) \cdot 1  \\ \text{Simply:}  \\  2x(x - 1) - (x + 1) = (x + 1)(x - 1)$$
3. Expand: $$2x^2 - 2x - x -1 = x^2 -1 \\  \text{Simply:}  \\ 2x^2 -3x -1 = x^2 -1$$
4. Rearrange into a quadratic equation: $$2x^2 - 3x - 1 - x^2 + 1 = 0 \\ x^2 - 3x = 0$$
5. Factorize: $$x(x - 3) = 0$$
6. Solve for $x$: $$x = 0  \text{or}  x = 3$$
7. Check the restrictions: $x \not = -1,1$. Both solutions are valid.

Final Answer: $$x = 0 \text{or}  x = 3$$