Solving Ax + By = Cx + Dy: A=C & B=D?

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SUMMARY

The discussion centers on the equation Ax + By = Cx + Dy and the conditions under which A = C and B = D can be concluded. It is established that if "=" denotes "identically equal to" (≡), then A must equal C and B must equal D for all values of x and y. Conversely, if "=" is interpreted as "equals" in a specific instance, A and C may not necessarily be equal, as demonstrated by a counterexample where A = 5, C = 3, B = 1, and D = 4, yielding a true statement without equality between A and C or B and D.

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  • Understanding of algebraic equations and identities
  • Familiarity with the concept of "identically equal to" (≡)
  • Basic knowledge of variable manipulation in equations
  • Experience with counterexamples in mathematical proofs
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  • Study the properties of algebraic identities and their implications
  • Learn about the differences between "equals" and "identically equal to" in mathematics
  • Explore the use of counterexamples in mathematical reasoning
  • Investigate the implications of variable dependencies in algebraic equations
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Homework Statement


I'm in a calculus course, but this isn't really a calculus question. I was wondering if:

Ax + By = Cx + Dy

Could I say A = C and B = D?

Homework Equations


None.

The Attempt at a Solution


None.
 
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Voivode said:

Homework Statement


I'm in a calculus course, but this isn't really a calculus question. I was wondering if:

Ax + By = Cx + Dy

Could I say A = C and B = D?
It depends on what "=" means. In the first equation, if "=" means [itex]\equiv[/itex] (identically equal to), then yes, A = C and B = D. Here, "identically equal to" means for any choice of x and y.

On the other hand, if "=" means just plain old "equals", then no.

Here's an example. Suppose x = 3 and y = 2. Your equation is equivalent to (A - C)x = (D - B)y, so if A = 5 and C = 3, and B = 1 and D = 4, then we have (5 - 3)3 = (4 - 1)2, or 6 = 6, a true statement, but A and C aren't equal, nor are B and D.
 
Last edited:
That makes sense. Thanks.
 

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