Chapter 11.1, Problem 26AYU

To determine

To find: The solution of the given system of equations.

Answer to Problem 26AYU

Solution:

No solution, Inconsistent system.

Explanation of Solution

Given:

  $x-y=5$

  $-3x+3y=2$

Calculation:

The system of equations is,

  $x-y=5$

  $-3x+3y=2$

Multiplying both sides of the first equation by -3 we get,

  $-3\times (x-y)=(-3)\times 5$

  $-3x+3y=-15$

The second equation is $-3x+3y=2$

So now the equations are,

  $-3x+3y=-15$

  $-3x+3y=2$

The value of the coefficient determinant is,

  $=\left|\begin{array}{ll}-3\hfill & 3\hfill \\ -3\hfill & 3\hfill \end{array}\right|=(-3)\times 3-3\times (-3)=-9+9=0$

The two equations $-3x+3y=-15$ and $-3x+3y=2$ both have same LHS but the RHS is different, which cannot be possible.

Therefore the given system of equations has no solution or is inconsistent.

The graph is as follows:

  

The graph of the given system of equations represents two parallel and non-intersecting lines.