Question

Solve the system of equations: 2x+y+2z=16 x+3y−z=37 4x−y+3z=4

Answer 1

Hello there. To solve this question, we have to remember some properties about system of equations.

Given the following system of linear equations:

$\begin{cases}2x+y+2z=16\\x+3y-z=37\\4x-y+3z=4\end{cases}$

We'll rewrite it in the form of augmented matrix:

$\begin{bmatrix}\begin{array}{ccc|c}2 & 1 & 2 & 16 \\ 1 & 3 & -1 & 37 \\ 4 & -1 & 3 & 4\end{array} \end{bmatrix}$

Now, we want to find the row-echelon form of this system, that is, we'll multiply the equations by a factor in order to find an equivalent system and we can solve it for each of the variables separately.

Divide the first equation by a factor of 2, getting:

$\begin{bmatrix}\begin{array}{ccc|c}1 & \frac{1}{2} & 1 & 8 \\ 1 & 3 & -1 & 37 \\ 4 & -1 & 3 & 4\end{array}\end{bmatrix}$

Multiply the first equation by a factor of (-1) and add it to the second equation. This happens because we chose the factor under the first term as the pivot for the second line.

$\begin{bmatrix}\begin{array}{ccc|c}1 & \frac{1}{2} & 1 & 8 \\ 0 & \frac{5}{2} & -2 & 29 \\ 4 & -1 & 3 & 4\end{array}\end{bmatrix}$

Next, multiply the first equation by a factor of (-4) and add it to the third equation

$\begin{bmatrix}\begin{array}{ccc|c}1 & \frac{1}{2} & 1 & 8 \\ 0 & \frac{5}{2} & -2 & 29 \\ 0 & -3 & -1 & -28\end{array}\end{bmatrix}$

Now, we move to the second equation. Choosing the element a22 as the pivot, we have to multiply the second equation by a factor of 6/5 and add it to the third equation:

$\begin{bmatrix}\begin{array}{ccc|c}1 & \frac{1}{2} & 1 & 8 \\ 0 & \frac{5}{2} & -2 & 29 \\ 0 & 0 & -\frac{17}{5} & \frac{34}{5}\end{array}\end{bmatrix}$

Back into the system of equations form, we have

$\begin{cases}x+\frac{1}{2}y+z=8 \\ \\ \frac{5}{2}y-2z=29 \\ \\ -\frac{17}{5}z=\frac{34}{5}\end{cases}$

Solving for z, we get

$-\dfrac{17}{5}z=\dfrac{34}{5}\Rightarrow z=-2$

Plugging it into the second equation, we get

$\begin{gathered} \dfrac{5}{2}y-2\cdot(-2)=29 \\ \\ \dfrac{5}{2}y=25 \\ \\ \Rightarrow y=10 \end{gathered}$

Finally, plugging it in the first equation, we find

$\begin{gathered} x+\frac{1}{2}\cdot10-2=8 \\ \\ \Rightarrow x=5 \end{gathered}$

Therefore the solution to this system of equations is:

$S=\{(x,\,y,\,z)\in\mathbb{R}^3|(x,\,y,\,z)=(5,\,10,\,-2)\}$

This is the final answer to this question.