Question

if x= 4a² + b² - 6ab, y= 3b² - 2a² + 8ab and z= 6a² + 8b² - 6ab, find : x-y-z
Plz don't answer like this
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Answer 1

Step-by-step explanation:

If x = 4a2 + b2 – 6ab. y = 3b2 – 2a2 + 8ab and z = 6a2 + 8b2 – 6ab find :

(i) x + y + z (ii) x – y – z

fundamental concepts (including fundamental operations)

x = 4a2 + b2 – 6ab

y = 3b2 – 2a2 + 8ab

z = 6a2 + 8b2 – 6ab

(i) x + y + z = 4a2 + b2 – 6ab + 3b2 – 2a2 + 8ab + 6a2 + 8b2 – 6ab

= 4a2 – 2a2 + 6a2 + b2 + 3b2 + 8b2 – 6ab + 8ab – 6ab

= 10a2 – 2a2 + 12b2 – 12ab + 8ab

= 8a2 + 12b2 – 4ab

(ii) x – y – z = (4a2 + b2 – 6ab) – (3b2 – 2a2 + 8ab) – (6a2 + 8b2 – 6ab)

= 4a2 + b2 – 6ab – 3b2 + 2a2 – 8ab – 6a2 – 8b2 + 6ab

= 4a2 + 2a2 – 6a2 + b2 – 3b2 – 8b2 – 6ab – 8ab + 6ab

= 6a2 – 6a2 + b2 – 11b2 – 14ab + 6ab

= -10b2 – 8ab

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Answer 2

Answer:

Step-by-step explanation:

(4a² +b²-6ab)-(3b²-2a²+8ab)-(6a²+ 8b²-6ab)

4a² +b²-6ab-3b²+2a²-8ab-6a²- 8b²+6ab)

4a²+2a²-6a² +b²-3b²-8b²-6ab-8ab+6ab

-10b²-8ab