$\text{a)}$  

$\{\begin{array}{l}2{\left(x-y\right)}^2-3\left(x+1\right)=2\left(x-y\right)\left(x+y\right)+4y\left(1-x+y\right)\\ \frac{x+y}{4}=1\mid \cdot 4\end{array}$  

$\{\begin{array}{l}2\left(x^2-2xy+y^2\right)-3x-3=2\left(x^2-y^2\right)+4y-4xy+4y^2\\ x+y=4\end{array}$  

$\{\begin{array}{l}2x^2-4xy+2y^2-3x-3=2x^2-2y^2+4y-4xy+4y^2\\ x+y=4\end{array}$  

$\{\begin{array}{l}-3x-3=4y\\ x+y=4\end{array}$  

$\{\begin{array}{l}-3x-4y=3\\ y=-x+4\end{array}$  

Podstawiamy y=-x+4 do pierwszego równania.

$\{\begin{array}{l}-3x-4\left(-x+4\right)=3\\ y=-x+4\end{array}$  

$\{\begin{array}{l}-3x+4x-16=3\\ y=-x+4\end{array}$  

$\{\begin{array}{l}x=19\\ y=-x+4\end{array}$  

Podstawiamy x=19 do drugiego równania.

$\{\begin{array}{l}x=19\\ y=-19+4\end{array}$  

$\{\begin{array}{l}x=19\\ y=-15\end{array}$  

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$\text{b)}$  

$\{\begin{array}{l}\frac{x+y}{2}-\frac{x-y-2}{4}=1\frac{1}{4}y\\ 2\left(x-3\right)+3\left(x+y\right)=-3\end{array}$  

$\{\begin{array}{l}\frac{x+y}{2}-\frac{x-y-2}{4}=\frac{5}{4}y\mid \cdot 4\\ 2\left(x-3\right)+3\left(x+y\right)=-3\end{array}$   

$\{\begin{array}{l}2\left(x+y\right)-\left(x-y-2\right)=5y\\ 2x-6+3x+3y=-3\end{array}$  

$\{\begin{array}{l}2x+2y-x+y+2=5y\\ 5x+3y=3\end{array}$  

$\{\begin{array}{l}x=2y-2\\ 5x+3y=3\end{array}$  

Podstawiamy x=2y-2 do drugiego równania.

$\{\begin{array}{l}x=2y-2\\ 5\left(2y-2\right)+3y=3\end{array}$  

$\{\begin{array}{l}x=2y-2\\ 10y-10+3y=3\end{array}$  

$\{\begin{array}{l}x=2y-2\\ 13y=13\mid :13\end{array}$  

$\{\begin{array}{l}x=2y-2\\ y=1\end{array}$  

Podstawiamy y=1 do pierwszego równania.

$\{\begin{array}{l}x=2\cdot 1-2\\ y=1\end{array}$  

$\{\begin{array}{l}x=2-2\\ y=1\end{array}$  

$\{\begin{array}{l}x=0\\ y=1\end{array}$