-2 3/4 as an improper fraction is -11/4. We need to ensure that the 5 fractions use +, -, * and / between them to get this answer, without using fractions whose denominator is 4. Multiplication first: a/b*b/c gives the result a/c, because the b's cancel out, and a/c*c/d gives a/d; but a/c*c/d is the same as a/c divided by d/c. So we need to find 3 fractions made up of a, b, c and d to give us the useful result a/d. The final result must be negative and neither multiplication nor division gives negative results if a, b, c and d are positive, so we'll use the minus operator in front of the result a/d. So part of our answer is going to be ...-a/b*b/c÷d/c. We just need to ensure that the remaining part, represented by e/f+g/h, is smaller than this so that the final expression comes to -11/4.
The final result will look like e/f+g/h-a/b*b/c÷d/c. Let's take the first part: e/f+g/h=(he+fg)/fh, and (he+fg)/fh-a/d=-11/4. Combining these fractions we get (d(he+fg)-afh)/dfh=-11/4. Looks like a tall order, doesn't it? But we know that dfh must be a multiple of 4 so that we end up with 4 in the denominator. So the numerator must also be a multiple of 4 so that the final fraction cancels down, and we know that this multiple must not itself be a multiple of 4. Let's pick a composite number that isn't a multiple of 4: 35, for example. That means dfh=4*35 and that means we have 4*5*7. We can't use f and h as the 4 because that would give us a fraction with 4 as the denominator, so d=4. We can now put f=5 and h=7. Now we have (4(7e+5g)-35a)/140=-11/4. We can write this as 16(7e+5g)-140a=-1540 by cross-multiplying. How do we find a, e and g?
If we rewrite this: 16(7e+5g)=140a-1540, we can see that a>38 to make the right side positive, which it must be to find positive values for e and g. Furthermore, 140a-1540 must be divisible by 16 otherwise we can't find integer values for e and g. This can be simplified if we write only the remainders after dividing by 16: 12a-4. We need to find a so that 12a-4 is divisible by 16. If a=3 then 12a-4=32, which is divisible by 16. In fact, a can be 3, 19, 35, 51, etc. that is, any multiple of 16 plus 3. The lowest value greater than 38 is 51. This would make 140a-1540=5600 (=350*16). What about e and g? 7e+5g=350, or e=(5(70-g))/7. Clearly, if g is a multiple of 7 we can find e. So let g=49 (7^2, which has an exponent) then e=5*21/7=15.
To recap, we have a=51, d=4, e=15, f=5, g=49, h=7. Our expression becomes: 15/5+49/7-51/b*b/c÷4/c=-11/4.
Since they cancel out, we can make b and c anything we like. Let's pick c=111 and b=73
15/5+7^2/7-51/73*73/111÷4/111
Because of the choice we had in selecting the fractions we can find many more solutions, perhaps replacing the improper fractions by proper fractions and choosing fractions without common factors. But the question doesn't ask for that.
