For what values of 'p' and 'q',da following system of linear equations will hav infinite number of solutions?

Question

2x - (p - 4)y=2q + 14x - (p - 1)y=5q - 1.

Answer 1

Im assuming these are 2 different equations and that the + sign was a typo:

2x-(p-4)y=2q

14x-(p-1)y-5q-1

Lets set up an augmented matrix

2    -(p-4)   2q

14   -(p-1)  5q-1

Now we must put the matrix into echelon form by Subtracting 7 times the top row from the second row:

The top row remains the same and I'll expand the parenthesis

2    4-p             2q

0    (1-p)-7(4-p)  5q-1-14q

Further simplification yields

2    4-p     2q

0    6p-27  -9q-1

Now for this system to have infinite number of solution we must have only one pivot in the first row and 6p-27 must equal to 0

So p must equal 9/2

Now if -9q-1 does not equal to 0 than instead of infinitely many solutions you will have an undeterminite matrix

So -9q-1=0

q=-1/9

Answer 2

2x - (p-4)y = 2q ....... 14x - 7(p-4)y = 14q

14x - (p-1)y = 5q - 1

since 14 = 7*2 .. . .

then

p - 1 = 7(p-4)

6p = 27

p = 9/2

and

14q = 5q - 1

9q = -1

q = -1/9

.. . .. .