To solve by the equation system by substitution, first, write the third equation for one of the variables, for example, write it for x:
[tex]\begin{gathered} x-z=4 \\ x-z+z=4+z \\ x=4+z \end{gathered}[/tex]Replace the expression obtained in x in the first equation:
[tex]\begin{gathered} 2x+3y-4z=-3 \\ 2(4+z)+3y-4z=-3 \\ 2\cdot4+2\cdot z+3y-4z=-3 \\ 6+2z+3y-4z=-3 \\ 6+2z-4z+3y=-3 \\ 6-2z+3y=-3 \end{gathered}[/tex]Write the equation for y:
[tex]\begin{gathered} 6-2z+3y=-3 \\ 6-6-2z+3y=-3-6 \\ -2z+3y=-9 \\ -2z+2z+3y=-9+2z \\ 3y=-9+2z \\ \frac{3y}{3}=-\frac{9}{3}+\frac{2z}{3} \\ y=-3+\frac{2}{3}z \end{gathered}[/tex]Use the expressions obtained for x and y in the second equation:
[tex]\begin{gathered} -x-2y+3z=4 \\ -(4+z)-2(-3+\frac{2}{3}z)+3z=4 \end{gathered}[/tex]From this expression, you can calculate the value of z:
[tex]\begin{gathered} -4-z-\cdot2(-3)+(-2)(\frac{2}{3}z)+3z=4 \\ -4-z+6-\frac{4}{3}z+3z=4 \\ -z-\frac{4}{3}z+3z-4+6=4 \\ \frac{2}{3}z+2=4 \\ \frac{2}{3}z+2-2=4-2 \\ \frac{2}{3}z=2 \\ (\frac{2}{3}\cdot\frac{3}{2})z=2\cdot\frac{3}{2} \\ z=3 \end{gathered}[/tex]Using the value of z you can calculate the values of y and x
Start with y:
[tex]\begin{gathered} y=-3+\frac{2}{3}z \\ y=-3+\frac{2}{3}\cdot3 \\ y=-3+2 \\ y=-1 \end{gathered}[/tex]Next, calculate the value of x:
[tex]\begin{gathered} x=4+z \\ x=4+3 \\ x=7 \end{gathered}[/tex]The solution of the equation system is x=7, y=-1 and z=3
