September 2015 – Mathe Solutions

$\begin{equation*} \frac{3x+4}{x-5}-\frac{x-9}{x-7}=\frac{2x^2-13x+27}{x^2-12x+35} \end{equation*}$

Lösung

$\begin{eqnarray*} \frac{(3x+4)(x-7)-(x-9)(x-5)}{(x-5)(x-7)} & = & \frac{2x^2-13x+27}{(x-5)(x-7)} \\ 3x^2-21x+4x-28-\left(x^2-5x-9x+45\right) & = & 2x^2-13x+27 \\ 2x^2-3x-73 & = & 2x^2-13x+27 \\ 10x & = & 100 \end{eqnarray*}$

$\begin{equation*} \underline{\underline{x=10}} \end{equation*}$

Monat: September 2015

Dreiecksberechnungen

326c – FW Algebra

Lösung