Chapter 11, Problem 3CT

To determine

The solution of the system of linear equations $\{\begin{array}{l}x-y+2z=5\\ 3x+4y-z=-2\\ 5x+2y+3z=8\end{array}$ by substitution method or elimination method. If system has no solution, then state it is inconsistent.

Answer to Problem 3CT

Solution:

The solution of the system $\{\begin{array}{l}x-y+2z=5\\ 3x+4y-z=-2\\ 5x+2y+3z=8\end{array}$ is $\left\{\left(x,y,z\right)|x=-z+\frac{18}{7},y=z-\frac{17}{7},z\text{is}\text{a}\text{real}\text{number}\right\}$.

Explanation of Solution

Given information:

The system of linear equations $\{\begin{array}{l}x-y+2z=5\\ 3x+4y-z=-2\\ 5x+2y+3z=8\end{array}$.

Explanation:

Let’s solve the system of equation by elimination method, The given system of linear equations is

$x-y+2z=5$ … (1)

$3x+4y-z=-2$ … (2)

$5x+2y+3z=8$ ‥ (3)

This is the system of three linear equations and has three variables.

The reduced form of a system of equation of three variables into a system of two variables

To eliminate variable $y$, multiply equation (3) by $2$ and subtract from (2),

$\begin{array}{l}3x+4y-z=-2\\ -\underset{\_}{10x+4y+6z=16}\\ -7x+0y-7z=-18\end{array}$

Thus, $7x+7z=18$

Multiply equation (1) by $2$ and add with equation (3)

$\begin{array}{l}5x+2y+3z=8\\ +\underset{\_}{2x-2y+4z=10}\\ 7x+0y+7z=18\end{array}$

Thus, $7x+7z=18$

The equivalent system of equation becomes,

$x-y+2z=5$ …(3)

$7x+7z=18$ …(4)

$7x+7z=18$ … (5)

As equation (4) and (5) are equivalent so has infinitely many solutions. This is the consistent system with dependent variable.

To solve equation (4) for $x$

$\Rightarrow 7x=18-7z$

$\Rightarrow x=\frac{18}{7}-z$

Substitute the value of $x$ in equation (1) and to solve for $y$ in terms of $z$.

$\Rightarrow \left(\frac{18}{7}-z\right)-y+2z=5$

$\Rightarrow -y+z=5-\frac{18}{7}$

$\Rightarrow -y+z=\frac{17}{7}$

$\Rightarrow -y=-z+\frac{17}{7}$

$\Rightarrow y=z-\frac{17}{7}$

The solution of the system is $x=\frac{18}{7}-z$ and $y=z-\frac{17}{7}$, $z$ is a real number.

Therefore, the solution of the system $\{\begin{array}{l}x-y+2z=5\\ 3x+4y-z=-2\\ 5x+2y+3z=8\end{array}$ is $\left\{\left(x,y,z\right)|x=-z+\frac{18}{7},y=z-\frac{17}{7},z\text{is}\text{a}\text{real}\text{number}\right\}$.