how do you solve N+3/N+1=3/2?

Answer 1

cross multiply

I am assuming you mean (N+3) / (N+1) = 3/2

multiply both sides by 2
2(N+3) / (N+1) = 3

multiply both sides by (N+1)
2(N+3) = 3(N+1)

expand the brackets
2N + 6 = 3N + 3

Solve for N


If you actually mean

N + (3/N) + 1 = 3/2

This is a little more advanced. You would need to have

N^2 + 3 + N = 3N/2 {multiply both sides by N}

N^2 + N - 3N/2 + 3 = 0

N^2 - N/2 + 3 = 0

Use quadratic equation to find the roots

N = ((1/2) +/- SQRT(1/4 - 4.1.3)) / 2

giving you 2 imaginary roots

N = 1/4 +/- SQRT(11&3/4)/2

very messy :(

Answer 2

Assume question reads as:-
(n + 3) / (n + 1) = 3 / 2
2n + 6 = 3n + 3
n = 3

Answer 3

N+3/N+1=3/2
N(N+3/N+1)=N(3/2)
N^2+3+N=3/2N ; N is not 0
N^2-N/2+3=0
2N^2-N+6=0
N= [ -(-1)+ - sqrt( (-1)^2-4*2*6 ) ] /2*2 = [ 1+ - sqrt(-47) ] /4 = no answer

Answer 4

2n^2+6+2n=3n

2n^2-1n+6=0

...
reffering the question, the task is not :
(n+3)/N .... but actually n+3/n ....that in mathematics means : n+ (3/n) ...

If he discriminant ( sqrt(-47)) is <0,
the result of the above should be two complex numbers ...

(1+sqrt(-47))/4 and
(1-sqrt(-47))/4

Answer 5

2(N+3)=3(N+1) (Cross multiply)
2N+6=3N+3
3=N (Combining like terms on the same side)

Answer 6

mutliply both sides by (n+1)
so it becomes n+3 = 3/2(n+1)
then n+3 = 3/2n + 3/2
then subtract n from both sides and subtract 3/2 from both sides
1/2n = 1.5
n =3

Answer 7

very simple. cross mutiply it

2x (n+3) = 3x (n+1)
2n+ 6= 3n + 3
2n-3n = -6+3
-n=-3
n=3

thats the answer

Answer 8

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Answer 9

n+3/n+1=3/2

lcd=2(n+1)

2(n+3)=3(n+1)
2n+6=3n+3
2n-2n+6=3n-2n+3
6=n+3
3=n

Answer 10

Wrong question.