Question

If P= a^2-b^2+2ab, Q=a^2+4b^2 -6ab, R=b^2+b, S=a^2-4ab and T= -2a^2+b^2-ab+a. Find P+Q+R+S-T ni​

Answer 1

Step-by-step explanation:

Given:-

P= a^2-b^2+2ab, Q=a^2+4b^2 -6ab, R=b^2+b, S=a^2-4ab and

T= -2a^2+b^2-ab+a.

To find:-

Find P+Q+R+S-T ?

Solution:-

P= a^2-b^2+2ab

Q=a^2+4b^2 -6ab

R=b^2+b

S=a^2-4ab

T= -2a^2+b^2-ab+a

Now,

P+Q+R+S-T

= (a^2-b^2+2ab)+(a^2+4b^2 -6ab)+(b^2+b)+(a^2-4ab)-(-2a^2+b^2-ab+a)

=>a^2-b^2+2ab+a^2+4b^2 -6ab+b^2+b +a^2- 4ab +2a^2-b^2+ab-a

=>(a^2+a^2+a^2+2a^2)+(-b^2+4b^2+b^2-b^2) + (2ab-6ab-4ab+ab) +(a)+(b)

=>(5a^2)+(3b^2)+(-7ab)+(a)+(b)

=>5a^2+3b^2-7ab+a+b

P+Q+R+S-T = 5a^2+3b^2-7ab+a+b

Answer:-

The value of P+Q+R+S-T for the given problem is 5a^2+3b^2-7ab+a+b