Chapter 3, Problem 8E

Solution

To state: The system of equations as independent, dependent, or inconsistent. Then solve the independent systems by graphing.

  $\{\begin{array}{c}1+y=x\\ x+y=1\end{array}$

The resultant solution is $\left(1,0\right)$ .

Given information:

The given system is:

  $\{\begin{array}{c}1+y=x\\ x+y=1\end{array}$

Explanation:

For two equations $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ .

The system of equation will be independent if: $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

The system of equation will be dependent if: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

The system of equation will be inconsistent if: $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Now consider the given system:

  $\{\begin{array}{c}1+y=x\\ x+y=1\end{array}$

The equations can be written as:

  $\begin{array}{c}-x+y+1=0\\ x+y-1=0\end{array}$

Now check if the system of equations is independent, dependent, or inconsistent:

  $\frac{-1}{1}\ne \frac{1}{1}$

Therefore, the lines will intersect or the lines are independent.

Now consider the equation $1+y=x$ :

For $x=0$ ,

  $\begin{array}{c}1+y=0\\ y=-1\end{array}$

For $y=0$ ,

  $\begin{array}{c}1+0=x\\ 1=x\end{array}$

Now consider the equation $x+y=1$ :

For $x=0$ ,

  $\begin{array}{c}0+y=1\\ y=1\end{array}$

For $y=0$ ,

  $\begin{array}{c}x+0=1\\ x=1\end{array}$

Now graph each equation:

  

From the graph the solution is $\left(1,0\right)$ .