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Problem
Prove that, where a, b, c, d, e are real numbers and a <> 0, if ax + by = c has the same solution set as ax + dy = e, then they are the same equation.
Given Solution
If a <> 0 then solution set of the first equation is {(x,y) | x = (c - by)/a}. Taking y = 0 gives the solution (c/a, 0), and since the second equation is supposed to have the same solution set, substituting into it gives a(c/a) + d(0) = e, so c = e. Then taking y = 1 in x = (c - by)/a gives a((c - b)/a) + d = e, which gives b = d. Hence they are the same equation.
My Thoughts
I don't buy into the solution above because it assumes that one of the members of the solution set has y = 0 and that another has y = 1. And anyways, you can take any two-variable two-equation linear system (where the equations aren't equal) and solve to get the solution set. Is this problem bogus or what?
Prove that, where a, b, c, d, e are real numbers and a <> 0, if ax + by = c has the same solution set as ax + dy = e, then they are the same equation.
Given Solution
If a <> 0 then solution set of the first equation is {(x,y) | x = (c - by)/a}. Taking y = 0 gives the solution (c/a, 0), and since the second equation is supposed to have the same solution set, substituting into it gives a(c/a) + d(0) = e, so c = e. Then taking y = 1 in x = (c - by)/a gives a((c - b)/a) + d = e, which gives b = d. Hence they are the same equation.
My Thoughts
I don't buy into the solution above because it assumes that one of the members of the solution set has y = 0 and that another has y = 1. And anyways, you can take any two-variable two-equation linear system (where the equations aren't equal) and solve to get the solution set. Is this problem bogus or what?