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2x^2x\ 3x y^2x\ 4x^3y^5 is equal to: 24 ...

`2x^2x\ 3x y^2x\ 4x^3y^5` is equal to: `24 x^6y^6` (b) `24 x^6y^7` `24 x^7y^6` (d) `24 x^7y^7`

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Subtract: (i) 5a + 7b - 2c from 3a - 7b + 4c (ii) a - 2b - 3c from -2a + 5b - 4c (iii) 5x^(2) - 3xy + y^(2) from 7x^(2) - 2xy - 4y^(2) (iv) 6x^(3) - 7x^(2) + 5x - 3 from 4 - 5x + 6x^(2) - 8x^(3) (v) x^(3) + 2x^(2) y + 6xy^(2) - y^(3) from y^(3) - 3xy^(2) - 4x^(2) (vi) -11 x^(2) y^(2) + 7xy - 6 from 9x^(2) y^(2) - 6xy + 9 (vii) -2a + b + 6d from 5a - 2b - 3c

Solve 4x+3y=24 ; 3y-2x=6

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RD SHARMA-INTRODUCTION TO ALGEBRA-All Questions
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