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Solve the following 4 equations. (2 marks each) 

(n+2)! 
-------- = 42 
n! 


(n+1)! 
-------- = 6 
(n-1)! 


(n-15)! 
--------- = 1/12 
(n-14)! 


(n+1)! 
-------------- = 20 
(n-2)!(n-1) 

asked Nov 12, 2014 in ALGEBRA 2 by anonymous

4 Answers

0 votes

(1)

The function is

image

image

image

n²+3n+2 = 42

n²+3n-40 = 0

n² + 8n - 5n - 40 = 0

n(n + 8) - 5(n + 8) = 0

(n + 8)(n - 5) = 0

n = -8, 5

As n cannot be a negative integer, n = 5.

Therefore the value of n is 5.

answered Nov 12, 2014 by Lucy Mentor
0 votes

(2)

The function is

image

image

image

image

n² + 3n - 2n - 6 = 0

n(n+3)-2(n+3) = 0

(n+3)(n-2) = 0

n = -3 and 2

We know that n cannot be a negative integer then n = 2

Therefore the value of n is 2.

answered Nov 12, 2014 by Lucy Mentor
0 votes

(3)

The function is

image

image

n - 15 = 12

n = 15 + 12

n = 27.

Therefore the value of n is 27.

answered Nov 12, 2014 by Lucy Mentor
0 votes

(4)

The Function is

image

image

image

image

image

n(n + 5) -4(n+5) = 20

(n+5)(n-4) = 20

n = -5 and 4

As n cannot be a negative integer, n = 4.

Therefore the value of n is 4.

answered Nov 12, 2014 by Lucy Mentor

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