Question

Triangular Systems

Use back-substitution to solve the triangular system

$\left\{\begin{array}{l}3x-3y+z=0\\ y+4z=10\\ z=3\end{array}\right.$

Short answer

Hence we get the solution of the triangular system as

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

Step-by-step solution

Step 1. Given information

We are given the set of linear equations that is-

$\left\{\begin{array}{l}3x-3y+z=0\\ y+4z=10\\ z=3\end{array}\right.$

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:

$\begin{array}{l}3x-3y+z=0\\ y+4z=10\\ z=3\end{array}$

From the above system of equations, we observe that $z=3$, substituting it in the equation $y+4z=10$, we get

$\begin{array}{l}y+4\left(3\right)=10\\ y+12=10\\ y=10-12\\ y=-2\end{array}$

Now substituting the values of $y,z$in the equation $3x-3x+z=0$, we get

$\begin{array}{l}3x-3(-2)+3=0\\ 3x=-9\\ x=\frac{-9}{3}\\ x=-3\end{array}$

So, we solve the triangular system.