Given:
The system of equations is,
[tex]\begin{gathered} 3x-y-z=2.\text{ . .. . . .(1)} \\ x+y+2z=4\text{ . . . .. . (2)} \\ 2x-y+3z=9\text{ . . . . . . .(3)} \end{gathered}[/tex]The objective is to solve the equations using the elimination method.
Explanation:
Consider the equations (1) and (2).
[tex]\begin{gathered} 3x-y-z=2 \\ \frac{x+y+2z=4}{4x+z=6} \\ \ldots\ldots\ldots.(4)\text{ } \end{gathered}[/tex]Now, consider the equations (2) and (3).
[tex]\begin{gathered} x+y+2z=4 \\ \frac{2x-y+3z=9}{3x+5z=13} \\ \ldots\ldots\text{ . . . .. (5)} \end{gathered}[/tex]On multiplying the equation (4) with (-5),
[tex]\begin{gathered} -5\lbrack4x+z=6\rbrack \\ -20x-5z=-30\text{ . . . . . .(6)} \end{gathered}[/tex]To find x :
On solving the equations (5) and (6),
[tex]\begin{gathered} 3x+5z=13 \\ \frac{-20x-5z=-30}{-17x=-17} \\ x=\frac{-17}{-17} \\ x=1 \end{gathered}[/tex]To find z :
Substitute the value of x in equation (6),
[tex]\begin{gathered} -20(1)-5z=-30 \\ -5z=-30+20 \\ -5z=-10 \\ z=\frac{-10}{-5} \\ z=2 \end{gathered}[/tex]To find y :
Now, substitute the values of x and z in equation (2).
[tex]\begin{gathered} x+y+2z=4 \\ 1+y+2(2)=4 \\ y=4-1-4 \\ y=-1 \end{gathered}[/tex]Hence, the value of x is 1, y is -1 and z is 2.
