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Equations with fractions

Summary and examples

Roughly speaking an equation with fractions is an equation with the independent variable (for example x) appearing in the denominator or in both the numerator and denominator.

In a number of examples we clarify how to solve equations with fractions.

Example 1

Solve:

\displaystyle\frac{1}{x}=\displaystyle\frac{x-1}{x+1}

The equation is only defined for denominators unequal to 0:

x\neq0 and x\neq{-1}

Essential in this first method is that fractions can only be added (or subtracted) if they have equal denominators. That is why we make both denominators equal:

\displaystyle\frac{1}{x}.\displaystyle\frac{x+1}{x+1}=\displaystyle\frac{x-1}{x+1}.\displaystyle\frac{x}{x}

resulting in the equation:

\displaystyle\frac{x+1-x(x-1)}{x(x+1)}=0

or:

\displaystyle\frac{-x^2+2x+1}{x(x+1)}=0

or:

\displaystyle\frac{x^2-2x-1}{x(x+1)}=0

A fraction is equal to 0 if the numerator is equal to 0, thus:

x^2-2x-1=0

This equation is solved by applying the abc-formula:

x_1=\displaystyle\frac{2+\sqrt{8}}{2}=1+\sqrt{2}

x_2=\displaystyle\frac{2-\sqrt{8}}{2}=1-\sqrt{2}

A second method (cross-multiplication) is straightforward. Cross-multiplication means that the product of 1 and x+1 is made equal to the product x and x-1, thus:

1.(x+1)=x(x-1)

resulting in the same equation.

Usually cross-multiplication is preferred because it is a faster method. However, this method fails if we have inequalities with fractions. We come back to this in detail in Inequalities with fractions.

Example 2

Solve:

\displaystyle\frac{x+1}{x-1}=\displaystyle\frac{x-1}{x+1}

The equation is only defined for denominators unequal to 0:

x\neq1 and x\neq{-1}

Cross-multiplication yields:

(x+1)^2=(x-1)^2

or:

x^2+2x+1=x^2-2x+1

4x=0

so:

x=0

Example 3

Solve:

\displaystyle\frac{e^{x}-1}{e^{x}}=\displaystyle\frac{1}{e^{x}+1}

Both denominators are not equal to 0 for all values of x, and thus the equation is defined for alle values of x.

Cross-multiplication yields:

(e^{x}-1)(e^{x}+1)=e^{x}

e^{2x}-1=e^{x}

e^{2x}-e^{x}-1=0

This is a quadratic equation in e^{x} and the solution is (using the abc-formula):

e^{x_{1,2}}=\displaystyle\frac{1\pm\sqrt{5}}{2}

Just one of the two values is positive and thus valid since an exponential function cannot be negative. So:

e^{x}=\displaystyle\frac{1+\sqrt{5}}{2}

and thus, after having applied the natural logarithm:

x=\ln{(\displaystyle\frac{1+\sqrt{5}}{2})}

See also Logarithmic functions and graphs

Example 4

Solve:

\displaystyle\frac{3}{x-3}-\displaystyle\frac{2}{x+3}=\displaystyle\frac{9}{x^2-9}

The equation is only valid if the denominators are unequal to 0:

x\neq3 en x\neq{-3}

In order to solve this equation we have to make the denominators in the left-hand side equal:

\displaystyle\frac{3}{x-3}.\displaystyle\frac{x+3}{x+3}-\displaystyle\frac{2}{x+3}.\displaystyle\frac{x-3}{x-3}=\displaystyle\frac{9}{x^2-9}

\displaystyle\frac{3(x+3)-2(x-3)}{(x+3)(x-3)}=\displaystyle\frac{9}{x^2-9}

\displaystyle\frac{x+15}{x^2-9}=\displaystyle\frac{9}{x^2-9}

and thus:

x=-6

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