I’m fine with basic rearrangements like turning y = ax + b into x = (y – b)/a. I get confused when the variable I want to make the subject shows up both in the numerator and the denominator, and I’m not sure how to treat steps that involve multiplying or dividing by expressions that might be zero. For example, if I try to make x the subject in y = (ax + b)/(cx + d), I can rearrange to something like x(yc – a) = b – yd. At that point, do I have to split into cases for yc – a = 0, and how do I keep track of the original condition cx + d ≠ 0 while still aiming for an equivalent formula for x?
I hit the same issue with T = L/(1 – rL) when solving for L. Multiplying by (1 – rL) seems to assume it’s nonzero, but that’s also a domain restriction of the original equation. Later I want to divide by something like (1 + Tr). Is there a clean checklist for carrying domain restrictions through these steps so the final formula is genuinely equivalent? A brief outline using either example would help me see the logic.
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3 Responses
Quick checklist I use: 1) write the “no-go” denominators in the margin first, 2) cross-multiply, 3) collect the thing you want, 4) divide by what’s in front of it, 5) at the end, make sure you didn’t divide by zero anywhere. Think of denominators like a pothole in the road: you don’t fix it by driving into it faster; you mark it (≠ 0) and drive around. For y = (ax + b)/(cx + d), start with cx + d ≠ 0. Cross-multiply: y(cx + d) = ax + b, so x(yc − a) = b − yd, and if yc − a ≠ 0 then x = (b − yd)/(yc − a). If yc − a = 0, you’re typically stuck (division by zero), though there’s a weird corner where also yd = b and then every x with cx + d ≠ 0 works. As for the original cx + d ≠ 0, plugging the solution in gives cx + d = (cb − da)/(yc − a). Since we already need yc − a ≠ 0, it’s “enough” that cb − da ≠ 0; if cb − da = 0 you’ve basically got a constant fraction, so either nothing or everything works depending on y. In practice I don’t split cases unless yc = a pops up.
For T = L/(1 − rL), same playbook. First note 1 − rL ≠ 0 (that was already true in the original). Multiply: T(1 − rL) = L, so T = L + TrL = L(1 + Tr). If 1 + Tr ≠ 0, then L = T/(1 + Tr). Nice bonus: the old restriction auto-checks out because 1 − rL = 1/(1 + Tr), so it’s nonzero whenever 1 + Tr is. If 1 + Tr = 0, that would force −1 = 0 when you plug back, so there’s no solution in that case. That’s the whole logic: write the “≠ 0”s first, only divide by things you’ve declared nonzero, and if a denominator you need to divide by can be zero, do a quick side-case check-most of the time it either kills all solutions or gives that “everything works” scenario.
Totally feel you on the “denominator booby-trap” feeling! My mental checklist goes like this: first, note the denominator’s don’t-touch values; second, cross-multiply to clear fractions; third, gather the x terms; fourth, divide by the coefficient of x, and finally remember to exclude any values you divided by that could be zero. For y = (a x + b)/(c x + d), I’d write y(cx + d) = ax + b, then (yc − a)x = b − yd, so x = (b − yd)/(yc − a). Here I’d say cx + d ≠ 0 from the start, and also yc − a ≠ 0 so that last division makes sense; if yc − a = 0 then the x sort of “vanishes,” which I usually treat as giving no solution for x (I’m sweeping the oddball special case where b − yd might also be 0 under the rug, oops). A quick numeric peek: take a=2, b=3, c=1, d=4, y=5; then x = (3 − 5·4)/(5·1 − 2) = (3 − 20)/(5 − 2) = −17/3, and indeed the original denominator 1·x + 4 is −17/3 + 4 = −5/3, not zero, so we’re safe. For T = L/(1 − rL), multiply to get T(1 − rL) = L, then T − TrL = L, so T = L(1 + Tr) and L = T/(1 + Tr); I like to say the old restriction 1 − rL ≠ 0 “turns into” 1 + Tr ≠ 0 after solving, so you just exclude Tr = −1 at the end. If you want a calmer walk-through of this kind of rearranging-with-restrictions, Khan Academy’s rational equations overview is handy: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:rational-expressions-equations.
Checklist: you may multiply by any denominator because any genuine solution already makes it nonzero; when you later divide by a factor, record it as “≠ 0” and separately check the case where it equals 0.
Example: from T = L/(1 − rL), multiply to T(1 − rL) = L (valid since solutions have 1 − rL ≠ 0), rearrange to L(1 + Tr) = T, so L = T/(1 + Tr) with 1 + Tr ≠ 0; if 1 + Tr = 0 (i.e., T = −1/r) the equation gives a contradiction, and for L = T/(1 + Tr) we also have 1 − rL = 1/(1 + Tr) ≠ 0.