Add the following:

$(i) 5 a^{2} b^{2} + 10 a b - 12 a^{2} + 15 b - 20$ and $34 a b + 24 a^{2} + 22 b + 24$

$(i i) a b c + 6 b c d + a b d - 9 a b + 4 b c - 9 a - 43$ and $4 a c d + 10 a c - 54 b - 32 c$

$(i i i) 5 b c + 9 a d - 3 c - 12$ and $24 - 5 c - 5 b + 2 a b$

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Solution

$(i) 5 a^{2} b^{2} + 10 a b - 12 a^{2} + 15 b - 20 + 34 a b + 24 a^{2} + 22 b + 24$

$= 5 a^{2} b^{2} + 44 a b + 12 a^{2} + 37 b + 4$

$(i i) a b c + 6 b c d + a b d - 9 a b + 4 b c - 9 a - 43 + 4 a c d + 10 a c - 54 b - 32 c$

$= a b c + 6 b c d + a b d - 9 a b + 4 b c - 9 a - 43 + 4 a c d + 10 a c - 54 b - 32 c$

$(i i i) 5 b c + 9 a d - 3 c - 12 + 24 - 5 c - 5 b + 2 a b$

$= 5 b c + 9 a d - 8 c + 12 - 5 b + 2 a b$

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Similar questions

Add the following:

$(i) 5 a^{2} b^{2} + 10 a b - 12 a^{2} + 15 b - 20$ and $34 a b + 24 a^{2} + 22 b + 24$

$(i i) a b c + 6 b c d + a b d - 9 a b + 4 b c - 9 a - 43$ and $4 a c d + 10 a c - 54 b - 32 c$

$(i i i) 5 b c + 9 a d - 3 c - 12$ and $24 - 5 c - 5 b + 2 a b$ [4 MARKS]

Add and express the sum as mixed fraction: (i) $\frac{- 12}{5} a n d \frac{43}{10}$ (ii) $\frac{24}{7} a n d \frac{- 11}{4}$ (iii) $\frac{- 31}{6} a n d \frac{- 27}{8}$ (iv) $\frac{101}{6} a n d \frac{7}{8}$

Q. If P = 7a - [6b - {9a - (5c + 12b) + 5c - 5 (3a - 5b)}]
and Q = 6a - [15b - {8a - (4c + 11b) + 4c}].
then (16P + 7Q) is equal to:

Q. Factorize:

12 a^{2} + 60 a b + 75 b^{2}

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Add the following: (i) 5a2b2+10ab−12a2+15b−20 and 34ab+24a2+22b+24 (ii) abc+6bcd+abd−9ab+4bc−9a−43 and 4acd+10ac−54b−32c (iii) 5bc+9ad−3c−12 and 24−5c−5b+2ab

(i) 5a2b2+10ab−12a2+15b−20+34ab+24a2+22b+24 ​ =5a2b2+44ab+12a2+37b+4 (ii) abc+6bcd+abd−9ab+4bc−9a−43+4acd+10ac−54b−32c =abc+6bcd+abd−9ab+4bc−9a−43+4acd+10ac−54b−32c (iii) 5bc+9ad−3c−12+24−5c−5b+2ab =5bc+9ad−8c+12−5b+2ab

Add the following:

$(i) 5 a^{2} b^{2} + 10 a b - 12 a^{2} + 15 b - 20$ and $34 a b + 24 a^{2} + 22 b + 24$

$(i i) a b c + 6 b c d + a b d - 9 a b + 4 b c - 9 a - 43$ and $4 a c d + 10 a c - 54 b - 32 c$

$(i i i) 5 b c + 9 a d - 3 c - 12$ and $24 - 5 c - 5 b + 2 a b$