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Solving Equations Involving Fractions with Linear Denominators

⭐ Higher Tier Content

Some equations contain fractions with linear expressions in the denominator. Solving these equations often leads to a linear equation or a quadratic equation after the fractions have been removed.

 

A clear and careful method is needed to avoid errors.

 

 

Understanding the Strategy

When an equation contains fractions with algebraic denominators, the first step is to clear the fractions. This is done by multiplying every term in the equation by the lowest common denominator.

 

This removes the fractions and produces an equation that can then be solved using standard methods.

 

 

Solving an Equation Leading to a Linear Equation

Consider the equation:

$$
\frac{x}{x + 2} = 3
$$

 

Multiply both sides by \( x + 2 \):

$$
x = 3(x + 2)
$$

 

Expand the bracket:

$$
x = 3x + 6
$$

 

Collect like terms:

$$
-2x = 6
$$

 

Divide by \( -2 \):

$$
x = -3
$$

 

Check the restriction. The denominator \( x + 2 \) cannot be zero, so \( x \neq -2 \).

Since \( x = -3 \) is allowed, it is a valid solution.

 

 

Solving an Equation Leading to a Quadratic Equation

Consider the equation:

$$
\frac{1}{x} + \frac{2}{x + 1} = 3
$$

 

The common denominator is \( x(x + 1) \). Multiply every term by this expression:

$$
(x + 1) + 2x = 3x(x + 1)
$$

 

Simplify the left side:

$$
3x + 1 = 3x^2 + 3x
$$

 

Rearrange so the equation equals zero:

$$
3x^2 + 3x - 3x - 1 = 0
$$

$$
3x^2 - 1 = 0
$$

 

Solve the quadratic:

$$
3x^2 = 1
$$

$$
x^2 = \frac{1}{3}
$$

$$
x = \pm \sqrt{\frac{1}{3}}
$$

 

Check restrictions. The denominators give \( x \neq 0 \) and \( x \neq -1 \).

Both solutions satisfy these conditions, so both are valid.

 

 

Equations with Brackets in the Denominator

Consider:

$$
\frac{2}{x - 1} = \frac{x}{2}
$$

 

The common denominator is \( 2(x - 1) \). Multiply both sides by this:

$$
4 = x(x - 1)
$$

 

Expand:

$$
4 = x^2 - x
$$

 

Rearrange:

$$
x^2 - x - 4 = 0
$$

 

Factorising does not work easily here, so the solutions are:

$$
x = \frac{1 \pm \sqrt{17}}{2}
$$

 

Check the restriction \( x \neq 1 \).

Both values are valid, so both solutions are accepted.

 

 

Checking for Invalid Solutions

When solving equations with algebraic denominators, always check for excluded values. Any value that makes a denominator equal to zero must be rejected, even if it satisfies the final equation.

 

 

Key Points to Remember

Multiply every term by the lowest common denominator to clear fractions.
Simplify carefully and rearrange the equation to equal zero if needed.
Solve the resulting linear or quadratic equation.
Always check solutions against the original denominators.

 

Solving equations involving fractions with linear denominators requires careful algebra, but following a clear method ensures accurate and valid solutions.