We can start by deciding what our solution will be. Let's use x=1, y=2, and z=0. Now, since we know their value beforehand we can set a mathematical expression involving them on the left-hand side of an equation, and then use the numerical result as the right-hand side to set up an equation.
The third equation is 3x -4y +2z = - 5. Now we have our system of three equations.
2x + 3y - 2z =8 & (I) 2x -2y + 2z = - 2 & (II) 3x -4y + 2z = - 5 & (III)
We can proceed to solve it using the Elimination Method. Notice that the coefficient of the variable z in Equation (I) is opposite to the ones in Equations (II) and (III). Then, we can add Equation (I) to Equation (II) and Equation (I) to Equation (III) to eliminate z and obtain a system of two equations and two variables.
2x + 3y -2z =8 & (I) 2x -2y +2z = - 2 & (II) 3x -4y +2z = - 5 & (III) 1.6cm
Eq.(II) + Eq. (I) 1.8cm Eq.(III) + Eq. (I) 0.1cm
2x -2y +2z = - 2 1.8cm 3x -4y +2z = - 5 0.18cm
+ 3.9cm + 2.7cm [-0.2em]
2x + 3y -2z =8 1.9cm2x + 3y -2z =8 0.3cm [-0.8em]
1.5cm
4x +y = 6 2cm 5x -y = 3 0.27cm
We obtain a system of two equations 0.8cm
with two variables 2.25cm
4x +y = 6 & (I) 5x -y = 3 & (II) 2.1cm
At this point we can solve is as we would do with any other systems of two equations. Notice that and the variable y has opposite coefficients in them, so we can use the Elimination Method again.
We find the expected result, x=1, y=2 and z=0. Notice that we could have used any value for the variables at the beginning and used any coefficients for the variables when constructing the equations. There are infinitely many possible solutions for this exercise, and this is only an example solution.
