When 5a-2 b-10b-2 c-2-21ab-32 is subtracted from 50-20b-2 c-2-4bc-ac-5ab- we get 18 -30b-2c-2 - 4bc - ac - 16ab - 5a-2b - 10b-2c-18 - 30b-2c-2 - 4bc - ac - 16ab - 5a-2b-20- 30b-2c-2 - 4bc - ac - 16ab - 5a-2b-22 - 30b-2c-2 - 4bc - ac -16ab - 5a-2b-

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Subtraction of Algebraic Expressions

When 5a^2 b +...

Question

When $5 a^{2} b + 10 b^{2} c^{2} - 21 a b + 32$ is subtracted from $50 - 20 b^{2} c^{2} - 4 b c - a c - 5 a b$ , we get

$18 + 30 b^{2} c^{2} - 4 b c - a c + 16 a b - 5 a^{2} b + 10 b^{2} c$

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$18 - 30 b^{2} c^{2} - 4 b c - a c + 16 a b - 5 a^{2} b$

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$20 - 30 b^{2} c^{2} - 4 b c - a c + 16 a b - 5 a^{2} b$

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$22 - 30 b^{2} c^{2} - 4 b c - a c - 16 a b - 5 a^{2} b$

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Solution

The correct option is B
$18 - 30 b^{2} c^{2} - 4 b c - a c + 16 a b - 5 a^{2} b$

$(50 - 20 b^{2} c^{2} - 4 b c - a c - 5 a b) - (32 + 10 b^{2} c^{2} - 21 a b + 5 a^{2} b)$

$50 - 20 b^{2} c^{2} - 4 b c - a c - 5 a b - 32 - 10 b^{2} c^{2} + 21 a b - 5 a^{2} b$

$= 18 - 30 b^{2} c^{2} - 4 b c - a c + 16 a b - 5 a^{2} b$

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ABC is a triangle, D is a point on AB such that $A D = \frac{1}{4} A B$ and E is a point on AC such that $A E = \frac{1}{4}$ AC. Prove that DE = $\frac{1}{4} B C$

When $5 a^{2} b + 10 b^{2} c^{2} - 21 a b + 32$ is subtracted from $50 - 20 b^{2} c^{2} - 4 b c - a c - 5 a b$ , we get

Q. Subtract:

(i) 5 a b + 4 a c - 6 a + 15 c - 15

from

10 a b + 22 b c - 13 a - 14 b - 16 c - 24

(i i) 5 a^{2} b + 10 b^{2} c^{2} - 10 b^{2} c - 21 a b + 32

from

50 - 20 b^{2} c^{2} - 4 b c - a c - 5 a b

Subtract:

$(i) 5 a b + 4 a c - 6 a + 15 c - 15$ from $10 a b + 22 b c - 13 a - 14 b - 16 c - 24$

$(i i) 5 a^{2} b + 10 b^{2} c^{2} - 10 b^{2} c - 21 a b + 32$ from $50 - 20 b^{2} c^{2} - 4 b c - a c - 5 a b$ . [2 MARKS]

When we add $5 a^{2} b + 6 a b + 7 b c + 54$ to $14 a b + 8 b c + 16$ , we will get:

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When 5a2b+10b2c2−21ab+32 is subtracted from 50−20b2c2−4bc−ac−5ab, we get

The correct option is B 18−30b2c2−4bc−ac+16ab−5a2b (50−20b2c2−4bc−ac−5ab)−(32+10b2c2−21ab+5a2b) 50−20b2c2−4bc−ac−5ab−32−10b2c2+21ab−5a2b =18−30b2c2−4bc−ac+16ab−5a2b

When $5 a^{2} b + 10 b^{2} c^{2} - 21 a b + 32$ is subtracted from $50 - 20 b^{2} c^{2} - 4 b c - a c - 5 a b$ , we get