Q 184. - Intermediate Algebra

Question

Solve the system of equations.

$\left\{\begin{array}{l}2x+3y+z=12\\ x+y+z=9\\ 3x+4y+2z=20\end{array}\right.$

Step-by-Step Solution

Answer

The system of equations has no solution.

Step 1. Given the information.

The system of equations is,

$\left\{\begin{array}{l}2x+3y+z=12........\left(1\right)\\ x+y+z=9.............\left(2\right)\\ 3x+4y+2z=20.....\left(3\right)\end{array}\right.$

Step 2. Eliminating x from equations (1) and (2).

Eliminating $x$ from equations (1) and (2).

$$\begin{gathered} 2x+3y+z=12........\left(1\right) \\ -2x-2y-2z=-18.......\left(2\right)\times -2 \end{gathered}$$

Solving the equations we get,

$y-z=-6........\left(4\right)$

Step 3. Eliminating x from equations (2) and (3).

Eliminating $x$ from equations (2) and (3).

$$\begin{gathered} -3x-3y-3z=-27........\left(2\right)\times -3 \\ 3x+4y+2z=20.............\left(3\right) \end{gathered}$$

Solving the equations we get,

$y-z=-7......\left(5\right)$

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

Solving equations (4) and (5),

$$\begin{gathered} y-z=-6........\left(4\right) \\ y-z=-7........\left(5\right) \end{gathered}$$

Subtracting the equations we get,

$0=1$

The system of equations has no solution.