Question

(4a²-7ab-4b²+1) - (3a²-6ab-3b²-1)=?
Can anyone pls send the procedure on how to solve this sum​

Answer 1

$Given \: (4a^{2}-7ab - 4b^{2}+1) -(3a^{2}-6ab-3b^{2}-1)$

$=4a^{2} -7ab - 4b^{2}+1 - 3a^{2}+6ab+3b^{2}+1$

/* Rearranging the terms, we get */

$=4a^{2} - 3a^{2} - 7ab + 6ab - 4b^{2} + 3b^{2} +1+1$

$= (4-3)a^{2}+(- 7 + 6)ab +(- 4+ 3)b^{2} +2$

$= a^{2} - ab - b^{2} +2$

Therefore.,

$\red{ (4a^{2} -7ab - 4b^{2}+1) -(3a^{2}-6ab-3b^{2}-1)}$

$\green {= a^{2} - ab - b^{2} +2}$

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Answer 2

Answer :

a^2 - ab - b^2 + 2

Step-by-step explanation:

(4a^2 - 7ab - 4b^2 +1) - (3a^2 - 6ab - 3b^2 - 1 )

= 4a^2 - 7ab - 4b^2 + 1 - 3a^2 + 6ab + 3b^2 + 1

= a^2 - ab - b^2 + 2 ≈ Answer