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lilavasa [31]
3 years ago
8

In the equation (x^2+y)^5, what is the coefficient of the term x^4y^3? what is the coefficient of the same term in the expansion

of (3x^2+y)^5?
Mathematics
1 answer:
lys-0071 [83]3 years ago
6 0

\displaystyle
(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k

<em>-------------------------------------------------------------</em>


\displaystyle&#10;(x^2+y)^n=\sum_{k=0}^n\binom{n}{k}x^{2n-2k}y^k\\&#10;n=5\\&#10;k=3\\\\\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdor5}{2}=10

<u>It's 10.</u>

----------------------------------------------------

\displaystyle&#10;(3x^2+y)^n=\sum_{k=0}^n\binom{n}{k}(3x)^{2n-2k}y^k=\sum_{k=0}^n\binom{n}{k}\cdot 3^{2n-2k}\cdot x^{2n-2k}y^k\\\\&#10;n=5\\&#10;k=4\\\\&#10;\binom{5}{3}\cdot3^{2\cdot5-2\cdot4}=10\cdot3^{2}=10\cdot9=90

<u>It's 90</u>

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-13x=90-2y <br> -6x=48-2y
Kisachek [45]

While it's pretty obvious to most of us that


-13x=90-2y

-6x=48-2y


is a system of linear equations, it'd be well to include that info plus the instructions "solve this system of linear equations."


Subtract the 2nd equation from the first:


-13x=90-2y

+6x=-48+2y

-----------------------

-7x = 42. Then x = -42/7, or x = 6.


Now subst. 6 for x in either one of the given equations. Suppose we use the 2nd equation:


-6x=48-2y


Then -6(6)=48-2y, or -36 = 48 - 2y, or 2y = 48+ 36 = 84. Then y = 42.


The solution is (6, 42).

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