a)

$\{\begin{array}{l}5x-3\frac{1}{2}y=60\\ 4x-3y=70\end{array}$

Obliczamy wyznaczniki W, Wx i Wy

$W=\left|\begin{array}{cc}5& -3\frac{1}{2}\\ 4& -3\end{array}\right|=5\cdot \left(-3\right)-\left(-3\frac{1}{2}\right)\cdot 4=-15-\left(-14\right)=-15+14=-1$

$W_x=\left|\begin{array}{cc}60& -3\frac{1}{2}\\ 70& -3\end{array}\right|=60\cdot \left(-3\right)-\left(-3\frac{1}{2}\right)\cdot 70=-180+\frac{7}{\sout{2}}\cdot \sout{70}=$

$=-180+7\cdot 35=-180+245=65$

$W_y=\left|\begin{array}{cc}5& 60\\ 4& 70\end{array}\right|=5\cdot 70-60\cdot 4=350-240=110$

W≠0 zatem układ równań jest oznaczony i rozwiązaniem jest para liczb:

$x=\frac{W_x}{W}$

$y=\frac{W_y}{W}$

$x=\frac{65}{-1}=-65$

$y=\frac{110}{-1}=-110$

 b)

$\{\begin{array}{l}\frac{6}{10}x+\frac{8}{10}y=5\\ 250x+300y=1\end{array}$

$W=\left|\left[\frac{6}{10},\frac{8}{10}\right],\left[250,300\right]\right|=\frac{6}{\sout{10}}\cdot \sout{300}-\frac{8}{\sout{10}}\cdot \sout{250}=6\cdot 30-8\cdot 25=180-200=-20$

$W_x=\left|\begin{array}{c}\frac{5,8}{10}\\ 1,300\end{array}\right|=5\cdot 300-\frac{8}{10}\cdot 1=1500-\frac{8}{10}=1499\frac{2}{10}$

$W_y=\mid \left[\frac{6}{10,5}\right],\left[250,1\right]=\frac{6}{10}\cdot 1-5\cdot 250=\frac{6}{10}-1250=-1249\frac{4}{10}$

$x=\frac{1499\frac{2}{10}}{-20}=\frac{14992}{10}\cdot \left(-\frac{1}{20}\right)=-\frac{\sout{14992}}{\sout{200}}=-\frac{7496}{100}=-74\frac{96}{100}$

$y=\frac{-1249\frac{4}{10}}{-20}=\frac{12494}{10}\cdot \frac{1}{20}=\frac{12494}{200}=\frac{6247}{100}=62\frac{47}{100}$