Question

for what value of p,q the system 2x+3y+5z=9 , 7x+3y+2z=8 , 2x+3y+pz=q has no solution many solutions​

Answer 1

Given : 2x+3y+5z=9 , 7x+3y+2z=8 , 2x+3y+pz=q

To Find : Value of p & q  for no solution and many solutions

Solution:

2x+3y+5z=9  

7x+3y+2z=8

2x+3y+pz=q

for not unique solution :

$\left|\begin{array}{ccc}2&3&5\\7&3&2\\2&3&p\end{array}\right|=0$

=> 2 (3p - 6) - 3(7p - 4) + 5(21 - 6) =0

=> 6p - 12  - 21p + 12 + 75 = 0

=> -15p  + 75 = 0

=> p = 5

p = 5 has no unique solution

hence either no solution  or many solutions

2x+3y+5z=9  

7x+3y+2z=8

2x+3y+5z=q

if q = 9

Then

2x+3y+5z=9  

7x+3y+2z=8

2 Equations in 3 variables hence Many solutions

if q ≠ 9

then 2x+3y+5z=9  

        2x+3y+5z ≠ 9

which is impossible hence no solution

p = 5 , q = 9  ( many solutions)

p = 5 , q ≠ 9  ( no solution )

p ≠ 5  ( unique solution ) , q ∈ R

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