Chapter 16.9, Problem 11WE

To determine

To calculate: The number of solution for the system of equations,

  $\begin{array}{c}2x-y=5\\ -2x+y=-1\end{array}$

Answer to Problem 11WE

The system of equations has no solution.

Explanation of Solution

Given information:

The system of equations are

  $\begin{array}{c}2x-y=5\\ -2x+y=-1\end{array}$

Formula used:

For system of equations

  $\begin{array}{c}ax+by=e\\ cx+dy=f\end{array}$

The determinant $D=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|\text{ and }{D}_y=\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|$ .

If $D\ne 0$ then the system has a unique solution.

If $D=0\text{ and }{D}_y\ne 0$ the system has no solution.

If $D=0\text{ and }{D}_y=0$ the system has infinite number of solutions.

Calculation:

Consider the system of equations are

  $\begin{array}{c}2x-y=5\\ -2x+y=-1\end{array}$

Recall that for system of equations

  $\begin{array}{c}ax+by=e\\ cx+dy=f\end{array}$

The determinant $D=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|\text{ and }{D}_y=\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|$ .

Apply it,

  $\begin{array}{c}D=\left|\begin{array}{cc}2& -1\\ -2& 1\end{array}\right|\\ =2-2\\ =0\end{array}$

And,

  $\begin{array}{c}{D}_y=\left|\begin{array}{cc}2& 5\\ -2& -1\end{array}\right|\\ =-2+10\\ =8\end{array}$

Recall that if $D\ne 0$ then the system has a unique solution.

If $D=0\text{ and }{D}_y\ne 0$ the system has no solution.

If $D=0\text{ and }{D}_y=0$ the system has infinite number of solutions.

Here, $D=0\text{ and }{D}_y\ne 0$ .

Thus, the system of equations has no solution.