Math, asked by sagarpakad78, 4 months ago

Add the following polynomials.
(i) x + y + xy, x – z + yx, and z + x + xz
(ii) 2x2y2– 3xy + 4, 5 + 7xy – 3x2y2, and 4x2y2 + 10xy
(iii) -3a2b2, (–5/2) a2b2, 4a2b2, and (⅔) a2b2​

Answers

Answered by Anonymous
6

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Answered by RvChaudharY50
14

Solution (i) :- x + y + xy, x – z + yx, and z + x + xz

→ (x + y + xy) + (x - z + yx) + (z + x + xz)

→ x + x + x + y + z + xy + xy + xz

→ 3x + y + z + xy + xz

→ 3x + xy + xz + y + z

x(3 + y + z) + (y + z) (Ans.)

Solution (ii) :- 2x²y²– 3xy + 4, 5 + 7xy – 3x²y², and 4x²y² + 10xy.

→ (2x²y²– 3xy + 4) + (5 + 7xy – 3x²y²) + (4x²y² + 10xy)

→ 2x²y² - 3x²y² + 4x²y² - 3xy + 7xy + 10xy + 4 + 5

→ 6x²y² - 3x²y² - 3xy + 17xy + 9

→ 3x²y² + 14xy + 9

xy(3xy + 14) + 9 (Ans.)

Solution (iii) :- -3a²b², (–5/2)a²b², 4a²b², and (⅔) a²b² .

→ -3a²b²+ (–5/2)a²b²+ 4a²b²+ (2/3)a²b²

→ a²b²[4 + (2/3) - (5/2) - 3]

→ a²b²[ (6*4 + 2*2 - 5*3 - 3*6)/6 ]

→ a²b²[(24 + 4 - 15 - 18)/6]

→ a²b²(28 - 33)/6

(-5/6)a²b² (Ans.)

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