# 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 a power function cannot be negative. So:

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

and thus, after having applied the natural logarithm:

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

##### 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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