Basic functions: Rational functions
Equations with rational functions
To solve equations with rational functions, use the following calculation rules for fractions.
Rules for solving equations with rational functions \[
\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.
\(x= -5\)
We have the following reduction: \[\begin{aligned}\frac{2}{x^2+x}&=\frac{1}{x^2+3x}&\\[0.25cm] 2\cdot (x^2+3x)&=(x^2+x)\cdot 1 &\blue{\text{crosswise product}}\\[0.25cm] 2x^2+6x&=x^2+x& \blue{\text{expansion of brackets}}\\[0.25cm] 2x^2-x^2+6x-x&=0& \blue{\text{collection of terms on the left-hand side}}\\[0.25cm] x^2+5x&=0& \blue{\text{simplification}}\\[0.25cm] x(x+5)&=0& \blue{\text{factorisation}}\\[0.25cm] x=0\quad\text{or}\quad x&= -5&\blue{\text{isolation of }x}\end{aligned}\] 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.
