{l}{a+b+c+d=20}{3a-2c=3}{b+d=6}{c+b=8} megoldása

$$\left\{ \begin{array} { l } { a + b + c + d = 20 } \\ { 3a -2c = 3 } \\ { b + d = 6} \\ { c + b = 8 } \end{array} \right.$$

Megoldás a(z) a, b, c, d változóra

$$a = \frac{31}{5} = 6\frac{1}{5} = 6,2$$

$$b=\frac{1}{5}=0,2$$

$$c = \frac{39}{5} = 7\frac{4}{5} = 7,8$$

$$d = \frac{29}{5} = 5\frac{4}{5} = 5,8$$