Question

-q+r =0
p+3q+ 3 r = 1
-p-2q-2 r =0​

Answer 1

Answer:

hope this answer helps

Answer 2

SOLUTION

-q+r=0.....(1)

p+3q+3r=1......(2)

TO PROVE: -p-2q-2r=0

PROOF,

Again,

On multiplying 2q with -1,

-p-3q-3r=-1

-p-2q-2r=-1+(q+r)

We have from 1, q=r

-p-2q-2r=-1+2q

0=-1+2q

=》2q=1=2r

So,

-p-2q-2r=0

=》-p-2=0

=》-1+3q+3r-2=0

=》 -3+3(q+r)=0

=》-3+6r=0

=》-3+3×2r=0

=》-3+3=0

=》0=0

Therefore LHS=RHS

Hence Proved...