Equations with rational functions

Basic functions: Rational functions

Equations with rational functions

To solve equations with rational functions, use the following calculation rules for fractions.

$\begin{array}{l l l} \dfrac{t}{n}=0 & \implies t=0 & \text{A fracton is 0 if the numerator is 0.}\\ \dfrac{t}{n}=p &\implies t=p\cdot n & \text{multipliction by on the left- and right-hand side $n$.}\\ \dfrac{t}{n}=\dfrac{p}{q} & \implies t\cdot q =p\cdot r & \text{crosswise multiplication.}\\ \dfrac{t}{n}=\dfrac{t}{p} & \implies t=0 \;\vee\; n=p &\\ \dfrac{t}{n}=\dfrac{p}{n} & \implies t=p & \end{array}$ Because of possible division by zero, you always have to check afterwards whether the algebraically found solutions of equations with rational functions are valid solutions of the original equation.

Determine the exact solutions of the equation $\frac{4}{x^2+x}=\frac{2}{x^2+3x}$

$x= -5$

We have the following reduction: $\begin{aligned}\frac{4}{x^2+x}&=\frac{2}{x^2+3x}&\\[0.25cm] 4\cdot (x^2+3x)&=(x^2+x)\cdot 2 &\blue{\text{crosswise product}}\\[0.25cm] 4x^2+12x&=2x^2+2x& \blue{\text{expansion of brackets}}\\[0.25cm] 4x^2-2x^2+12x-2x&=0& \blue{\text{collection of terms on the left-hand side}}\\[0.25cm] 2x^2+10x&=0& \blue{\text{simplification}}\\[0.25cm] x(2x+10)&=0& \blue{\text{factorisation}}\\[0.25cm] x=0\quad\text{or}\quad x&= -5&\blue{\text{isolation of }x}\end{aligned}$ But $x=0$ is not a solution because substitution in the left-hand side of the original equation is not possible because of division by zero. The remaining solution is $x= -5$ because in this case substition in the original equation does not lead to division by zero.

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