Atrue mathematician would know this problem. so if you know the magic square(you can look it up)i need each of a 3 by 3 square to have one of these each number in it: 5,6,7,8,9,10,11,12, and 13. you can only use one number once. but each row, column and from diagonal must all equal up to 27.

Attached is one version.

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The key is that 9 must go in the middle.

If you know how to work a 3 by 3 magic square using the digits 1 to 9, you can make any 3 by 3 magic square starting with any number.

Starting Square

8  1   6

3  5  7

4  9  2

This magic square adds up to 15. That is 3 * the center number.

That is not the magic square you want, but it can be made into yours.

Just add 4 to every number in the base magic square.

12 5  10

7   9  11

8  13  5

And the sum is 3 * 9 = 27 in all the rows columns and diagonals, just what you requested. What happens when you have 0 in the center?

That means you take away 5 from every number in the base square.

3    - 4      1

-2     0     2

-1      4     -3

It still works. Every row column and diagonal add to 0.  

where are the options?

step-by-step explanation:

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x