Svolgimento:
[math]\begin{cases} xy=100 \\ (\frac{1}{2})x+2y=20 \ \end{cases}[/math]
Calcoliamo, nella prima equazione la [math]y[/math]
in funzione della [math]x[/math]
, e sostituiamo nella seconda [math]{(y=(100)/x),((\frac{1}{2})x+2((100)/x)=20):}[/math]
cioè [math]{(y=5,(x=20):}[/math]
![Sistemi: [math] {\left\lbrace\begin{matrix}\frac{4}{{3}}{x}+\frac{1}{{2}}{y}=\frac{15}{{4}}\\{y}-\frac{x}{{2}}-{1}=\frac{3}{{4}}\end{matrix}\right.} [/math]](https://cdn.skuola.net/shared/thumb/159x141/news_foto/images/stories/esercizi/sist_e5.jpg)
![Sistemi: Risolvere graficamente il sistema [math] {\left\lbrace\begin{matrix}{2}{x}-{3}{y}={5}\\{x}+{y}={2}{x}-{1}\end{matrix}\right.} [/math]](https://cdn.skuola.net/shared/thumb/159x141/news_foto/images/stories/esercizi/sist_e7.jpg)
![Sistemi:[math]{\left\lbrace\begin{matrix}{x}+{y}=\frac{a}{{b}}{\left({x}+{y}\right)}\\{2}\frac{{{a}^{2}+{b}^{2}}}{{{a}{b}}}-\frac{{{x}-{y}}}{{a}}=\frac{{{x}+{y}}}{{b}}\end{matrix}\right.}[/math]](https://cdn.skuola.net/shared/thumb/159x141/news_foto/images/stories/esercizi/sist_e3.jpg)
![Sistemi: [math]{\left\lbrace\begin{matrix}{\left({a}-{b}\right)}{x}+{\left({a}+{b}\right)}{y}={a}+{b}\\\frac{x}{{{a}+{b}}}-\frac{y}{{{a}-{b}}}=\frac{1}{{{a}+{b}}}\end{matrix}\right.}[/math]](https://cdn.skuola.net/shared/thumb/159x141/news_foto/images/stories/esercizi/sist_e9.jpg)
