Problem 6 - College Algebra Essentials

Question

Solve each system. $$\left\\{\begin{array}{rr} 2 x+y-2 z= & -1 \\ 3 x-3 y-z= & 5 \\ x-2 y+3 z= & 6 \end{array}\right.$$

Step-by-Step Solution

Answer

The given system of equations has no solution due to a contradiction in our calculations.

Step 1: Simplification

The second equation can be simplified by dividing all terms by -1. This results in the following system: \[ \left\{ \begin{array}{rr} 2x + y - 2z = & -1 \ 3x + 3y + z = & -5 \ x - 2y + 3z = & 6 \end{array} \right. \]

Step 2: Subtraction of the 1st and 2nd Equation

The 1st equation is subtracted from the 2nd, and this provides a new 2nd equation. This results in the system: \[ \left\{ \begin{array}{rr} 2x + y - 2z = & -1 \ x - 2z = & -4 \ x - 2y + 3z = & 6 \end{array} \right. \]

Step 3: Solve for \(x\)

By solving the modified 2nd equation, we obtain \(x = 4\).

Step 4: Substitute \(x\)

Substitute \(x = 4\) in the 1st and 3rd equation and solve for \(y\): \[ \left\{ \begin{array}{rr} y = & 2 \ y = & 3 \end{array} \right. \] This is a contradiction as \(y\) can't be 2 and 3 at the same time.

Step 5: Conclusion

This system of equations has no solution as we have a contradiction in Step 4 with respect to the value of \(y\).