Q. 373 - 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 z from the equations (1) and (2).

Eliminating $z$ from the equations (1) and (2).

$$\begin{gathered} 2x+3y+z=12........\left(1\right) \\ -x-y-z=-9.......\left(2\right)\times -1 \end{gathered}$$

Solving the equations, we get,

$x+2y=3.......\left(4\right)$

Step 3. Eliminating z from the equations (2) and (3).

Eliminating $z$ from the equations (2) and (3).

$$\begin{gathered} -2x-2y-2z=-18..........\left(2\right)\times -2 \\ 3x+4y+2z=20...............\left(3\right) \end{gathered}$$

Solving the equations, we get,

$x+2y=2......\left(5\right)$

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

Solving equations (4) and (5),

$$\begin{gathered} x+2y=3..........\left(4\right) \\ x+2y=2..........\left(5\right) \end{gathered}$$

Subtracting the equations, we get,

$0=1$

The system of equations has no solution.