1.

The system of linear equations 2x+2y-3z=1 4x+4y+z=2 6x+6y-z=3 has

Answer»

Given :

2x+2Y-3z=1

4x+4y+z=2

6x+6y-z=3

To Find: Solution

Solution:

2x+2y-3z=1    Eq1

4x+4y+z=2    Eq2

6x+6y-z=3     Eq3

2 * Eq1    - Eq2

=> -7z = 0  => z = 0

3* Eq1    -  Eq3

=> -8z = 0  =>  z  = 0

After substituting z = 0

2x + 2y  = 1

4x + 4y  = 2 => 2x + 2y = 1

6x + 6y = 3  => 2x + 2y = 1

Hence z = 0  and  infinite solutions for x and y  

Infinite solutions  and z = 0

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