Sistema 6

$\bigg \{ \begin{array}{ll} 2x-y=3 \\ 4x+\frac 1 2 y=1 \end{array}$

$\bigg \{ \begin{array}{ll} 2x-y=3 \\ \frac {8x+y} 2 =\frac 2 2 \end{array}$

$\bigg \{ \begin{array}{ll} 2x-y=3 \\ 8x+y = 2 \end{array}$

Troviamo $y$ nella seconda equazione, e poi andiamola a sostituire nella prima.

$\bigg \{ \begin{array}{ll} 2x-y=3 \\ y = 2-8x \end{array}$

$\bigg \{ \begin{array}{ll} 2x-(2-8x)=3 \\ y = 2-8x \end{array}$

$\bigg \{ \begin{array}{ll} 2x-2+8x=3 \\ y = 2-8x \end{array}$

$\bigg \{ \begin{array}{ll} 10x=3+2 \\ y = 2-8x \end{array}$

$\bigg \{ \begin{array}{ll} x=\frac {5}{10}=\frac 1 2 \\ y=2-8x \end{array}$

$\bigg \{ \begin{array}{ll} x= \frac 1 2 \\ y= 2-8(\frac1 2) \end{array}$

$\bigg \{ \begin{array}{ll} x= \frac 1 2 \\ y= 2-4=-2 \end{array}$

Troviamo $y$ in entrambe le equazioni, così da risolvere poi un’equazione di primo grado con incognita $x$ e teniamo solo la seconda.

$\bigg \{ \begin{array}{ll} y=2x-3 \\ y = 2-8x \end{array}$

$\bigg \{ \begin{array}{ll} 2x-3=2-8x \\ y = 2-8x \end{array}$

$\bigg \{ \begin{array}{ll} 2x+8x=2+3 \\ y = 2-8x \end{array}$

$\bigg \{ \begin{array}{ll} 10x=5 \\ y = 2-8x \end{array}$

$\bigg \{ \begin{array}{ll} x=\frac {5}{10}=\frac 1 2 \\ y=2-8x \end{array}$

$\bigg \{ \begin{array}{ll} x= \frac 1 2 \\ y= 2-8(\frac1 2) \end{array}$

$\bigg \{ \begin{array}{ll} x= \frac 1 2 \\ y= 2-4=-2 \end{array}$

Eseguiamo la somma membro a membro, lasciando la seconda equazione così com’è.

$\bigg \{ \begin{array}{ll} 2x+8x=2+3 \\ y = 2-8x \end{array}$

$\bigg \{ \begin{array}{ll} 10x=5 \\ y = 2-8x \end{array}$

$\bigg \{ \begin{array}{ll} x=\frac {5}{10}=\frac 1 2 \\ y=2-8x \end{array}$

$\bigg \{ \begin{array}{ll} x= \frac 1 2 \\ y= 2-8(\frac1 2) \end{array}$

$\bigg \{ \begin{array}{ll} x= \frac 1 2 \\ y= 2-4=-2 \end{array}$

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