Question

Triangular Systems

Use back-substitution to solve the triangular system

$\left\{\begin{array}{l}x-2y+3z=10\\ 2y-z=2\\ 3z=12\end{array}\right.$

Short answer

Hence we get the solution of the triangular system as

$\begin{array}{l}x=-6\\ y=3\\ z=4\end{array}$

Step-by-step solution

Step 1. Given information

We are given the set of linear equations that is-

$\left\{\begin{array}{l}x-2y+3z=10\\ 2y-z=2\\ 3z=12\end{array}\right.$

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:

$\begin{array}{l}x-2y+3z=0\\ 2y-z=2\\ 3z=12\end{array}$

Since we have

$\begin{array}{l}3z=12\\ z=4\end{array}$

Now we are substituting the value of z in equation $2y-z=2$, we get

$\begin{array}{l}2y-4=2\\ y=\frac{2+4}{2}\\ y=\frac{6}{2}\\ y=3\end{array}$

We will substitute the value of $y,z$in equation $x-2y+3z=0$

$\begin{array}{l}x-2\left(3\right)+3\left(4\right)=0\\ x-6+12=0\\ x+6=0\\ x=-6\end{array}$

So, we solve the triangular system.