To find the integer, you must cross multiply, group the same terms, then divide both side by 2:
Solution:
4/(x-3)= 2/3
12 = 2x-6
18=2x
x= 9
2000, you just need to put three zeros at the end of the number.
x | y
——-
-10|-18
-9 |-17
-8 |-16
-7 |-15
-6 |-14
-5 |-13
-4 |-12
-3 |-11
-2|-10
-1 | -9
0 | -8
1 | -7
2 |-6
3 | -5
4 | -4
5 | -3
6 | -2
7 | -1
8 | 0
9 | 1
10| 2
We can rewrite the expression under the radical as
then taking the fourth root, we get
![\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cleft%28%5Cdfrac32a%5E2b%5E3c%5E4%5Cright%29%5E4%7D%3D%5Cleft%7C%5Cdfrac32a%5E2b%5E3c%5E4%5Cright%7C)
Why the absolute value? It's for the same reason that
since both 
and 
return the same number , and 
captures both possibilities. From here, we have
The absolute values disappear on all but the 
term because all of 
, 
and 
are positive, while 
could potentially be negative. So we end up with
The first one means what is the smallest number that goes into 12 which would be 4. The second one mean what does not go into 4 and 6 so that would be 8 because 6 is not a multiple of 8.The third one is either 1 or 2 not sure. And the 4th one is little complicated you might want to look up video on prime factorization that can really help you .
