Al hallar la primera incógnita sustituimos y obtenemos la segunda. En los ejemplos siguientes lo veremos más claro.
\(\bullet\begin{cases} 2x+3y=7\\ 5x-y=9\end{cases} \)
\(\begin{cases} y=\dfrac {7-2x} {3}\\ y=5x-9\end{cases} \)
\(\dfrac {7-2x} {3}=5x-9\)
\(7-2x=15x-27\)
\(\boxed{x=2}\)
Sustituimos
\(y=5\cdot2-9\)
\(\boxed{y=1}\)
\(\bullet\begin{cases} x-2y=5\\ 2x-3y=9\end{cases} \)
\(\begin{cases} x=5+2y\\ x=\dfrac {9+3y} {2}\end{cases} \)
\(5+2y=\dfrac {9+3y} {2}\)
\(10+4y=9+3y\)
\(\boxed{y=-1}\)
Sustituimos
\(y=5-2\cdot1\)
\(\boxed{x=3}\)
\(\bullet \begin{cases} 2x+y=7\\ 5x-3y=1\end{cases} \)
\(\begin{cases} y=7-2\times \\ y=\dfrac {5x-1} {3}\end{cases} \)
\(7-2x=\dfrac {5x-1} {3}\)
\(21-6x=5x-1\)
\(\boxed{x=2}\)
Sustituimos
\(y=7-2\cdot2\)
\(\boxed{y=3}\)
\(\bullet\begin{cases} 5x-2y=5\\ 2x-3y=-9\end{cases} \)
\(\begin{cases} x=\dfrac {5+2y} {5}\\ x=\dfrac {-9+3y} {2}\end{cases} \)
\(\dfrac {5+2y} {5}=\dfrac {-9+3y} {2}\)
\(10+4y=-45+15y\)
\(\boxed{y=5}\)
Sustituimos
\(x=\dfrac {5+2\cdot 5} {5}\)
\(\boxed{x=3}\)
\(\bullet \begin{cases} 2x+5y=-2\\ 3x-2y=16\end{cases} \)
\(\begin{cases} x=\dfrac {-2-5y} {2}\\ x=\dfrac {16+2y} {3}\end{cases} \)
\(\dfrac {-2-5y} {2}=\dfrac {16+2y} {3}\)
\(-6-15y=32+4y\)
\(\boxed{y=-2}\)
Sustituimos
\(x=\dfrac {-2+5\cdot 2} {2}\)
\(\boxed{x=4}\)
