Question

Using identity solve (2a+3b)+(3a+2b)

Answer 1

Letting A=a⟹A/3=a/3.Letting A=a⟹A/3=a/3.

Then a/3=b/2Then a/3=b/2

⟹a3=b2⟹a3=b2

⟹a=3b2.⟹a=3b2.

So (2a+3b)/(3a−2b)So (2a+3b)/(3a−2b)

=2a+3b3a−2b=2a+3b3a−2b

=2(3b2)+3b3(3b2)−2b=2(3b2)+3b3(3b2)−2b

=3b+3b(9b2)−2b=3b+3b(9b2)−2b

=6b(9b2−4b2)=6b(9b2−4b2)

=6b(5b2)=6b(5b2)

=(12b2)(5b2)=(12b2)(5b2)

=12b5b=12b5b

=125, b≠0.

Answer 2

(2a+2b)+b+(2a+2b)+a
=2(2a+2b)+b+a
=2x2(a+b)+(a+b)
=(a+b)(4+1)
=5(a+b)