Question

subtract the sum of (6x^2+8x-9) and (-3+5x-3x^2
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Answer 1

$\huge\fcolorbox{red}{cyan}{Answer}$

1)As per question :-

$10-[(6x²+8x-9)+(-3+5x-3x²)] \\ => 10 - (6x²+8x-9-3+5x-3x²) \\ => 10 - (3x²+13x-12) \\ => 10 - 3x² - 13x + 12 \\ => \boxed{12-13x²-3x}$

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2) Let The Other Number be q

As per question :-

$(2x²-5x+7)+(q) = x²+9x-13 \\ => q = x²+9x-13-(2x²-5x+7) \\ => q = x²+9x-13-2x²+5x-7 \\ => \boxed{q = -x²+14x-20}$

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3)As per question :-

$[(x+3y-4z)+(4x-y+9z)+(-2x+3y-z)] - (2x-3y+4z) \\ => (x+3y-4z+4x-y+9z-2x+3y-z) - (2x-3y-4z) \\ => 3x+5y+4z - (2x-3y+4z) \\ =>3x+5y+4z-2x+3y-4z \\ => \boxed{x+8y}$

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4)Let the Other Number be q

As per question :-

$(3x³-2x²+3x-1)+(q) = x³+2x²-6x+5 \\ =>x³+2x²-6x+5-(3x³-2x²+3x-1) \\ =>q = x³+2x²-6x+5-3x³+2x²-3x+1 \\ => \boxed{q = -2x³+4x²-9x+6}$

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5)As per question :-

$\frac{-2x²}{3}+\frac{3xy}{2}+2y²-(q)=3x²-2xy-\frac{7y²}{2} \\ => \frac{-2x³}{3}+\frac{3xy}{2}+2y² = 3x²-2xy-\frac{7y²}{2}+q \\ => q = \frac{-2x²}{3}+\frac{3xy}{2}+2y² - (3x²-2xy-\frac{7y²}{2}) \\ =>\frac{-2x²}{3}+\frac{3xy}{2}+2y²-3x²+2xy+\frac{7y²}{2} \\ => \frac{-4x²+9xy+12y²-18x²+12xy+21y²}{6} \\ => \frac{-22x²+21xy+33y²}{6} \\ =>\boxed{\frac{11x²}{3}+\frac{7xy}{2}+\frac{11y²}{2}}$

Hope This Helped