# Find the solution of the system using Gauss elimination. \begin{alignat}{3} -8x + 2y &&=& -8,...

## Question:

Find the solution of the system using Gauss elimination.

{eq}\begin{alignat}{3} -8x + 2y &&=& -8, \\ 14x - 2y &&=& 8. \end{alignat} {/eq}

## Gauss Elimination Method:

Here, we have to solve the system of equations by using the Gauss Elimination method.

First, we rewrite the given system of equations into an augmented matrix. Then we use the Gauss elimination method to reduce the augmented matrix to a simpler form and then we find the solutions to the system.

## Answer and Explanation:

Given

{eq}\begin{alignat}{3} -8x + 2y &&=& -8, \\ 14x - 2y &&=& 8. \end{alignat} {/eq}

We have to find a solution to the system using Gauss elimination.

System of equations in matrix form is:

{eq}\begin{bmatrix} -8 & 2\\ 14 & -2 \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} -8\\ 8 \end{bmatrix} {/eq}

Its augmented matrix is:

{eq}\begin{align} C &=\left [ A: B \right ]\\ & \begin{bmatrix} -8 & 2 & : & -8\\ 14 & -2 & : & 8\\ \end{bmatrix}\\ \\ & R_1 \rightarrow R_1+R_2\\ \\ & \sim \begin{bmatrix} 6 & 0 & : & 0 \\ 14 & -2 & : & 8 \end{bmatrix}\\ \\ & R_2\rightarrow \frac{R_2}{2}\\ \\ & \sim \begin{bmatrix} 6 & 0 & : & 0\\ 7 & -1 & : & 4 \end{bmatrix} \end{align} {/eq}

which can be written as:

{eq}\begin{bmatrix} 6 & 0\\ 7 & -1 \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} 0\\ 4 \end{bmatrix}\\ \begin{bmatrix} 6x & \\ 7x-y & \end{bmatrix}=\begin{bmatrix} 0\\ 4 \end{bmatrix} {/eq}

On comparing:

{eq}\begin{align} 6x &=0\ & \enspace (A)\\ 7x-y &=4 \ & \enspace(B) \end{align} {/eq}

From {eq}(A) {/eq}

{eq}x=0 {/eq}

From {eq}(B) {/eq}

{eq}\begin{align} 7(0)-y &=4\\ y &=-4 \end{align} {/eq}

{eq}\color{blue}{\boxed{\therefore x=0, y=-4}} {/eq}

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from Algebra II Textbook

Chapter 10 / Lesson 6