Math, asked by sahusushila108, 5 hours ago

please help me sir or mam. please give full answer.​

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Answers

Answered by tennetiraj86
6

Step-by-step explanation:

Given :-

(2/3)x²-(4/5)xy³-(2/3) and (5/6)x²-(4/5)+(7/10)xy³

and (2/9)x²-(14/15)xy³+(7/15)

To find :-

Subtract the sum of (2/3)x²-(4/5)xy³-(2/3) and (5/6)x²-(4/5)+(7/10)xy³ from (2/9)x²-(14/15)xy³+(7/15)

Solution:-

Given expressions are :

(2/3)x²-(4/5)xy³-(2/3) and (5/6)x²-(4/5)+(7/10)xy³

and (2/9)x²-(14/15)xy³+(7/15)

Sum of (2/3)x²-(4/5)xy³-(2/3) and (5/6)x²-(4/5)+(7/10)xy³

=>[ (2/3)x²-(4/5)xy³-(2/3)] +[ (5/6)x²-(4/5)+(7/10)xy³]

=> [(2/3)x²+(5/6)x²]+[(-4/5)xy³+(7/10)xy³]+[(-2/3)-(4/5)]

=> [(2/3)+(5/6)]x² + [ (-4/5)+(7/10)]xy³+[(-2/3)-(4/5)]

=> [(4+5)/6]x²+[(-8+7)/10]xy³+[(-10-12)/15]

=> (9/6)x²+(-1/10)xy³+(-22/15)

On Subtracting (9/6)x²+(-1/10)xy³+(-22/15) from (2/9)x²-(14/15)xy³+(7/15)

=> [(2/9)x²-(14/15)xy³+(7/15)]-[(9/6)x²+(-1/10)xy³+(-22/15)]

=>(2/9)x²-(14/15)xy³+(7/15)-(9/6)x²+(1/10)xy³+(22/15)]

=> [(2/9)-(9/6)]x²+[(-14/15)+(1/10)]xy³+[(7/15)+(22/15)]

=> [(4-27)/18]x²+[(-28+3)/30]xy³+[(7+22)/15]

=> (-23/18)x²+(-25/30)xy³+(29/15)

=> (-23/18) x² -(5/6)xy³ + 29/15

Answer:-

Answer for the given problem is

(-23/18 )x² -(5/6) xy³ + 29/15

Used formulae:-

Change the signs of the terms in the second expression when subtract the second expresion from the first expression.

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