For three non-zero real numbers a, b, c if a/2 = b/3 = c/4 = (2a - 3b + 5c) /…

Concept: If a/p = b/q = c/r = k for non-zero real p, q, r, then a = pk, b = qk, and c = rk for a common ratio constant k, so any linear combination formed from a, b, c (for example m·a + n·b + o·c) equals k times the corresponding linear combination of p, q, r — the common ratio constant factors out of the whole expression.

Application:

1. Let the common ratio value be m, so a/2 = b/3 = c/4 = m, which gives a = 2m, b = 3m, and c = 4m.

2. Substitute these into the numerator: 2a - 3b + 5c = 2(2m) - 3(3m) + 5(4m).

3. Simplify each term and combine: 4m - 9m + 20m = 15m.

4. Since (2a - 3b + 5c)/k also equals m, set 15m/k = m; dividing both sides by the non-zero m gives 15/k = 1, so k = 15.

Cross-check: Taking m = 1 gives a = 2, b = 3, c = 4, so 2a - 3b + 5c = 4 - 9 + 20 = 15, and a/2 = 1. Requiring (2a - 3b + 5c)/k = 1 confirms k = 15, independently of the general algebra above.