Question

Triangular Systems

Use back-substitution to solve the triangular system

$\left\{\begin{array}{l}2x-y+6z=5\\ y+4z=0\\ -2z=1\end{array}\right.$

Short answer

Hence we get the solution of the triangular system as

$\begin{array}{l}x=5\\ y=2\\ z=\frac{-1}{2}\end{array}$

Step-by-step solution

Step 1. Given information

We are given the set of linear equations that is-

$\begin{array}{l}2x-y+6z=5\\ y+4z=0\\ -2z=1\end{array}$

Step 2. Concept used

The process of solving a linear system of equations that has been transformed into the row-echelon form or reduced row-echelon form is known as the back-substituting method.

We have to solve the given triangular system by using the back substitution method.

Step 3. Calculation

We have the triangular systems:

$\begin{array}{l}2x-y+6z=5\\ y+4z=0\\ -2z=1\end{array}$

Since we have

$\begin{array}{l}-2z=1\\ z=\frac{-1}{2}\end{array}$

Now we are substituting the value of z in equation $y+4z=0$, we get

$\begin{array}{l}y+4\left(\frac{-1}{2}\right)=0\\ y-2=0\\ y=2\end{array}$

We will substitute the value of $y,z$in equation $2x-y+6z=5$

We get,

$\begin{array}{l}2x-2+6\left(\frac{-1}{2}\right)=5\\ 2x=5+2+3\\ x=\frac{10}{2}\\ x=5\end{array}$

So, we solve the triangular system.