Using only the numbers 1-8 without repeating any of them, what is _+_=_-_=_-_=_-_=_? And the last number has to be the same as the first cause it's in a circle don't forget!!
To solve this equation, let's assign numbers from 1 to 8 to each of the variables. Since each number can only be used once, we will have:
A + B = C - D = E - F = G - H = A
Now, let's try to find the values for A, B, C, D, E, F, G, and H that satisfy the equation.
Given that the equation is in a circle, we can start with any variable and proceed clockwise or counterclockwise. Let's assign A = 1 and see if we can find a solution:
1 + B = C - D = E - F = G - H = 1
Next, we need to fill in the remaining numbers, making sure each number is used only once. Since B can't be 1, let's try B = 2:
1 + 2 = C - D = E - F = G - H = 1
Now, let's try different values for C, D, E, F, G, and H while keeping track of our progress:
1 + 2 = C - D = E - F = G - H = 1
3 = C - D
We can see that if C - D = 3, the only combination possible is C = 6 and D = 3:
1 + 2 = 6 - 3 = E - F = G - H = 1
Now, let's continue:
1 + 2 = 6 - 3 = E - F = G - H = 1
3 = E - F
Since E - F = 3, the only combination possible is E = 6 and F = 3 (we can't use 6 or 3 as they are already assigned to other variables):
1 + 2 = 6 - 3 = 6 - 3 = G - H = 1
Continuing:
1 + 2 = 6 - 3 = 6 - 3 = G - H = 1
3 = G - H
Since G - H = 3, the only combination possible is G = 6 and H = 3 (we can't use 6 or 3 as they are already assigned to other variables):
1 + 2 = 6 - 3 = 6 - 3 = 6 - 3 = 1
Now we have:
3 = A + B
3 = C - D
3 = E - F
3 = G - H
By assigning the values we found, we have:
3 = 1 + 2
3 = 6 - 3
3 = 6 - 3
3 = 6 - 3
Therefore, the equation becomes:
3 + 5 = 6 - 3 = 6 - 3 = 6 - 3 = 3
So, the solution to the equation is:
3 + 5 = 6 - 3 = 6 - 3 = 6 - 3 = 3