Q 188. - Intermediate Algebra

Question

Solve the system of equations.

$\left\{\begin{array}{l}x+y-3z=-1\\ y-z=0\\ -x+2y=1\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of equations is $(2z-1,z,z)$.

Step 1. Given the information

The system of equations is,

$\left\{\begin{array}{l}x+y-3z=-1........\left(1\right)\\ y-z=0............\left(2\right)\\ -x+2y=1..........\left(3\right)\end{array}\right.$

Step 2. Finding the value of y .

Simplifying equation (2), 

$$\begin{gathered} y-z=0 \\ y=z \end{gathered}$$

Step 3. Eliminating y from equation (1).

Substituting $y=z$ in the equation

$x+y-3z=-1$.

$$\begin{gathered} x+z-3z=-1 \\ x-2z=-1........\left(4\right) \end{gathered}$$

Step 4. Eliminating y from equation (3).

Substituting $y=z$ in the equation

$-x+2y=1$.

$-x+2z=1.......\left(5\right)$

Step 5. Solving equations (4) and (5).

Solving equations (4) and (5),

$$\begin{gathered} x-2z=-1......\left(4\right) \\ -x+2z=1......\left(5\right) \end{gathered}$$

Adding the equations we get,

$0=0$

The system of equation has infinitely many solutions.

Step 6. Showing how x is dependent on z

Simplifying the equation $x-2z=-1$,

$$\begin{gathered} x-2z=-1 \\ x=2z-1 \end{gathered}$$