Q 185. - Intermediate Algebra

Question

Solve the system of equations.

$\left\{\begin{array}{l}x+4y+z=-8\\ 4x-y+3z=9\\ 2x+7y+z=0\end{array}\right.$

Step-by-Step Solution

Answer

The system of equations has no real solution.

Step 1. Given the information.

The system of equations is,

$\left\{\begin{array}{l}x+4y+z=-8.........\left(1\right)\\ 4x-y+3z=9..........\left(2\right)\\ 2x+7y+z=0.........\left(3\right)\end{array}\right.$

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

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

$$\begin{gathered} -4x-16y-4z=32........\left(1\right)\times -4 \\ 4x-y+3z=9...............\left(2\right) \end{gathered}$$

Solving the equations we get,

$-17y-z=41......\left(4\right)$

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

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

$$\begin{gathered} 4x-y+3z=9...............\left(2\right) \\ -4x-14y-2z=0........\left(3\right)\times -2 \end{gathered}$$

Solving the equations we get,

$-15y+z=9........\left(5\right)$

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

Solving equations (4) and (5),

$$\begin{gathered} -17y-z=41........\left(4\right) \\ -15y+z=9.........\left(5\right) \end{gathered}$$

Solving the equations we get,

$$\begin{gathered} -32y=50 \\ y=-\frac{50}{32} \\ y=-\frac{25}{16} \end{gathered}$$

Here we cannot get any real solution.

The system of equations has no real solution.