Question

Triangular Systems

Use back-substitution to solve the triangular system

$\left\{\begin{array}{l}4x+3z=10\\ 2y-z=-6\\ \frac{1}{2}z=4\end{array}\right.$

Short answer

Hence we get the solution of the triangular system as

$\begin{array}{l}x=\frac{-7}{2}\\ y=1\\ z=8\end{array}$

Step-by-step solution

Step 1. Given information

We are given the set of linear equations that is-

$\left\{\begin{array}{l}4x+3z=10\\ 2y-z=-6\\ \frac{1}{2}z=4\end{array}\right.$

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

From the equation $\frac{1}{2}z=4$, we get

$\begin{array}{l}z=4\left(2\right)\\ z=8\end{array}$

We are substituting $z=8$ in equation $2y-z=-6$.

$\begin{array}{l}2y-8=-6\\ 2y=8-6\\ y=\frac{2}{2}\\ y=1\end{array}$

Now using the value of $z=8$in equation $4x+3z=10$, we get

$\begin{array}{l}4x+3\left(8\right)=10\\ 4x=10-24\\ x=\frac{-14}{2}\\ x=\frac{-7}{2}\end{array}$

So, we solve the triangular system.