What must be subtracted from 3 a2-6 a b-3 b2-1 to get 4 a2-7 a b-4 b2-1 -

Let the required number be x. (3a2 6ab 3b2 1) x=4a2 7ab 4b2+1 (3a2 6ab 3b2 1) (4a2 7ab 4b2+1)=x #160; #160; #160; #160;3a2 6ab 3b2 1 #160;4a2 7ab 4b ...

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Standard VIII

Mathematics

Addition and Subtraction of Algebraic Expression

What must be ...

Question

What must be subtracted from 3a² − 6ab − 3b² − 1 to get 4a² − 7ab − 4b² + 1?

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Solution

Let the required number be $x$ .
$(3 a^{2} - 6 a b - 3 b^{2} - 1) - x = 4 a^{2} - 7 a b - 4 b^{2} + 1$
$(3 a^{2} - 6 a b - 3 b^{2} - 1) - (4 a^{2} - 7 a b - 4 b^{2} + 1) = x$

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

∴ Required number = $- a^{2} + a b + b^{2} - 2$

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What must be subtracted from 3a2 − 6ab − 3b2 − 1 to get 4a2 − 7ab − 4b2 + 1?

Let the required number be x. (3a2-6ab-3b2-1)-x=4a2-7ab-4b2+1 (3a2-6ab-3b2-1)-(4a2-7ab-4b2+1)=x 3a2-6ab-3b2-1 4a2-7ab-4b2+1- + + - -a2+ ab+ b2-2 ∴ Required number = -a2+ab+b2-2

What must be subtracted from 3a² − 6ab − 3b² − 1 to get 4a² − 7ab − 4b² + 1?