Fractional First-Degree Equations

A Fractional First-Degree Equation is an equation containing algebraic fractions in which the unknown appears in the denominator and that can be reduced to integer first-degree equations.

To solve such a fractional equation we use the same rules derived from the principles of equivalence but with one extra precaution. We must ensure that the denominators containing the unknown are non-zero. For this reason we must impose Existence Conditions during the solution process.

In this lesson we will examine the solution scheme that lets us find the solutions of fractional first-degree equations.

Key Takeaways

Fractional first-degree equations can be reduced to integer first-degree equations;
To solve them, bring all algebraic fractions to a common denominator;
First, impose the existence conditions that keep denominators from vanishing;
After finding a solution, check that it satisfies the existence conditions.

Fractional First-Degree Equations

From the introductory lesson on equations we know that there are integer equations and fractional equations.

An equation is integer if the unknown never appears in a denominator. Conversely, in a fractional equation the unknown appears at least once in a denominator.

Here we focus on one-variable fractional equations of first degree—i.e. fractional equations that can be reduced to first-degree equations.

Definition

Fractional First-Degree Equation

A fractional first-degree equation is a one-variable fractional equation that can be reduced to an integer first-degree equation.

Examples of fractional first-degree equations:

\frac{x + 1}{x - 2} = \frac{3}{x - 2},

5 = \frac{3x}{x + 1}.

Existence Conditions

In general, the first step in solving a fractional first-degree equation is to remove the unknown from the denominators.

Consider

\frac{x}{x + 1} = 2.

To clear the denominator we apply the second principle of equivalence, multiplying both sides by the polynomial $x + 1$ :

\rightarrow\; (x + 1)\,\frac{x}{x + 1} = 2\,(x + 1).

We could now simplify the fraction in the left member, but only if we require that the polynomial be non-zero:

x + 1 \neq 0.

If $x + 1$ were zero, the division would be meaningless:

(x + 1) = 0 \;\Longrightarrow\; \frac{x + 1}{x + 1} = {\color{red}{\frac{0}{0}}}.

Thus the unknown must never take the value $-1$ :

x + 1 \neq 0 \;\Longrightarrow\; x \neq -1.

If a candidate solution violates this condition, the equation is impossible. Hence the condition is called an existence condition.

Continuing the solution, having imposed the condition we may simplify:

x \neq -1 \;\Longrightarrow\; \cancel{(x + 1)}\,\frac{x}{\cancel{x + 1}} = 2\,(x + 1).

We now have an integer first-degree equation:

\rightarrow\; x = 2\,(x + 1).

Solving it as a standard first-degree equation:

x = 2x + 2 \Longrightarrow\; x - 2x = 2 \Longrightarrow\; -x = 2 \Longrightarrow\; x = -2.

Finally, check the solution against the existence condition: since $-2 \neq -1$ , the solution is valid and the equation is solvable.

Definition

Existence Conditions for a Fractional First-Degree Equation

The Existence Conditions are the constraints imposed on the solution so that the equation is determined - namely, that all denominators containing the unknown are non-zero.

Solving a Fractional First-Degree Equation

Once the existence conditions are set, bring all algebraic fractions to a common denominator so the equation can be rewritten as an integer equation.

Consider the more complex example

\frac{3x}{x + 2} + \frac{2x}{x - 7} = \frac{5x + 6}{x + 2}.

Here we have three fractions and three denominators containing the unknown.

First, find the existence conditions by setting each denominator non-zero:

x + 2 \neq 0 \;\Longrightarrow\; x \neq -2, \qquad x - 7 \neq 0 \;\Longrightarrow\; x \neq 7.

Hence

x \neq -2 \,\wedge\, x \neq 7, \quad\text{i.e.}\quad x \in \mathbb{R}\setminus\{-2,\,7\}.

Now move every fraction to the common denominator $(x + 2)(x - 7)$ :

\frac{3x(x - 7)}{(x + 2)(x - 7)} \;+\; \frac{2x(x + 2)}{(x + 2)(x - 7)} \;=\; \frac{(5x + 6)(x - 7)}{(x + 2)(x - 7)}.

Because each fraction has the same denominator, apply the integer-coefficient reduction rule to eliminate the denominators:

3x(x - 7) + 2x(x + 2) = (5x + 6)(x - 7).

Expand and collect like terms:

3x^{2} - 21x + 2x^{2} + 4x = 5x^{2} - 35x + 6x - 42 \Longrightarrow\; 5x^{2} - 17x = 5x^{2} - 29x - 42.

Use the cancellation rule to remove $5x^{2}$ :

\cancel{5x^{2}} - 17x = \cancel{5x^{2}} - 29x - 42 \Longrightarrow\; 29x - 17x = -42 \Longrightarrow\; 12x = -42.

This is in the form $ax = b$ with $a \neq 0$ , so a solution exists:

x = -\frac{42}{12} = -\frac{7}{2}.

Finally, verify the solution against the existence conditions:

x \in \mathbb{R}\setminus\{-2,\,7\}.

Because $-\tfrac{7}{2}$ is neither −2 nor 7, it is admissible. The equation is therefore determinate.

Solution Scheme

Summarising, the solution scheme for a fractional first-degree equation is:

Hint

Determine the existence conditions by requiring that each denominator be non-zero.
Bring all algebraic fractions to a common denominator.
Multiply both sides by that denominator to obtain an integer equation.
Solve the resulting equation.
Check that the solution satisfies the existence conditions; otherwise the equation is impossible.

In summary

In this lesson we studied fractional first-degree equations, i.e. equations containing algebraic fractions in which the unknown appears in at least one denominator. They are first degree because they can be reduced to integer first-degree equations.

To solve them we impose existence conditions so denominators are non-zero, then convert to a common denominator and reduce to an integer equation. After solving, we must verify that the solution respects the existence conditions; if not, the equation is impossible.