Chapter 8, Problem 16RE

To determine

To solve: the system of linear equations, by Gaussian elimination with back-substitution.

Answer to Problem 16RE

The solutions are $x=-9$ and $y=-4$ .

Explanation of Solution

Given information:

Given system

  $\{\begin{array}{l}2x-5y=2\\ 3x-7y=1\end{array}$

Calculation:

Consider system of equations,

  $\{\begin{array}{l}2x-5y=2\\ 3x-7y=1\end{array}$

Write augmented matrix for equations,

  $\left[\begin{array}{cccc}2& -5& ⋮& 2\\ 3& -7& ⋮& 1\end{array}\right]$

Perform row operation $-R_2+R_1$

  $\left[\begin{array}{cccc}-1& 2& ⋮& 1\\ 3& -7& ⋮& 1\end{array}\right]$

Perform row operation $-R_1$

  $\left[\begin{array}{cccc}1& -2& ⋮& -1\\ 3& -7& ⋮& 1\end{array}\right]$

Perform row operation $-3R_1+R_2$

  $\left[\begin{array}{cccc}1& -2& ⋮& -1\\ 0& -1& ⋮& 4\end{array}\right]$

Perform row operation $-R_2$

  $\left[\begin{array}{cccc}1& -2& ⋮& -1\\ 0& -1& ⋮& -4\end{array}\right]$

This is row-echelon form and the corresponding system is,

  $\{\begin{array}{c}x-2y=-1\\ y=-4\end{array}$

Use back substitution,

Put $y=-4$ in $x-2y=-1$ and solve,

  $\begin{array}{c}x-2y=-1\\ x-2\left(-4\right)=-1\\ x+8=-1\\ x=-9\end{array}$

Hence, the solutions are $x=-9$ and $y=-4$ .