Here is the puzzle I will be going through, step by set to a solution.   The letters a-i are the missing numbers.

Instructions

Fill in the missing numbers.

Use the numbers 1 through 9 to complete the equations.

Each number is used only once.
Each row is a math equation. Work from left to right.
Each column is a math equation. Work from top to bottom.

Chart above will help you solve the puzzle

Look for a row or column that has the least number of possible solutions.
      a. All multiply or all divide are limiting.

C1 - a*d*g = 27 is a good place to start since it is all multiply and has a low product.

first factor 27. - ( 1*3*3*3 = 27 )  a,d, & g must be these factors.
since numbers 1-9 can only be used once, C1 must contain 1, 3, 9. in some order.
>    1*3*(3*3)=27  >   1*3*9=27
Using the chart to the right, mark off all numbers that a,d,g CAN NOT be, leaving open the numbers it can be (1, 3, 9). see chart>
Since we now know that 1, 3, or 9 will ALL be in C1, they can not be in any of the other rows or columns, so you can eliminate 1,3,6 from the others. see chart>

Since C3 is all multiply and has the next lowest product, lets try that one.

C3 - c*f*i = 112

Factor > 112 = 2*2*2*2*7. remember 1 is always a factor, but one has already been used in C1.
So: c,f,i can be 2,4,8, or 7 - Here 7 must be used, all factors must be used in some form.
C3 - must contain 2,8, & 7 because the only other way to get 112 by multiplying 3 numbers is: 4*4*7=112 and you can't use 4 twice

2*(2*2*2)*7=112 >  2*8*7=112
Now that we know all 3 number for C3 we can mark the chart as we did previously.

The next logical step seems to be C2, but we don't need to do it because from our chart we already know what they must be.
C2 - b*e*h = 120. from our chart b,e, & h are all left with 5,4, & 9. 4*5*6=120

Let look a R1.
R1 - (a-b)/c = -1, Remember these puzzles are worked from left to right and top to bottom, so a-b must be done 1st, hence the ( ). Not to be confused with a-(b/c) = -1.

There's some information in R1.
1. With multiply and divide, in order to get a negative answer (-1) one of the terms        must be negative. Since "c", can't = a negative number, then (a-b)<0.
     So: a<b  a can not be 9,  because there wont be any way "b" can be > than a.
2. Solve the equation for c and you get c = b-a.

a - is now down to being ether 1 or 3, so we can just substitute 1 & 3 for "a" with all possible "b" values (4,5,6) to see which ones work.
a=1: b=4 > 1-4 = c = 3 from the chart we see "c" can't be a 3, so that don't work
a=1: b=5 > c=4 that don't work, & a=1: b=6 > c=5 again that don't work.
"a" can not be 1 so "a" must be 3 the only number left and no other number can be a 3.
We now can solve for b & c by trying all possible answers for c = b-3.
b=4 > then c=1, don't work.  b=5 > c=2, that works.  b=6 > c=3, don't work.
b=5 & c=2, mark off on chart as before.

Try R3
R3 - g + h / i = 1, here i = g + h.

Try all possible answers for g & h to see if they work.
g = 1, h = 4 > i=5, don't work.  g=1, h=6 > i=7, it works.
g=9, h=4 > i=13, don't work.  g=9, h=6 > i=15 don't work.
g =1, h =6 & i=7  mark off on chart.

Your done. All other letters are left with only one possible solution.

The Formula method - good when you have lots of +'s & -'s

Start by looking for any 2 formulas, that when set = to each other, one variable cancels out.

Try (F2 = F5):
b - e - h + 11 = d - e + f  - 4
Set = to 0:
b - d - e + e - f - h + 15 = 0
F8:  b - d - f - h + 15 = 0

When eliminating variables don't use the same formula twice, because it will undo your previously eliminated variable.
Useing F8 = F3 we can eliminate the "f":
b - d - f - h + 15 = c - f + i
F9:  b - c - d - h - i + 15 = 0