B Certainty of being able to solve linear equation

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The discussion centers on the linear equation a + (2a - b)√2 = 4 + √2 and the confusion surrounding the number of solutions. Generally, with two variables and one equation, the system is considered underdetermined, leading to infinitely many solutions. However, participants clarify that if specific conditions are applied, such as restricting variables to integers, unique solutions can emerge. The equation can yield one value of 'a' for each 'b' when solved, indicating that the context and constraints significantly affect the solution set. Ultimately, the nature of the variables and any imposed conditions dictate whether solutions are unique or infinite.
Mr Davis 97
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I have the linear equation ##a + (2a - b) \sqrt{2} = 4 + \sqrt{2}##. I commonly hear that for linear equations, if we have two variables and one equation, then the system is undertermined and there are infinitely many solutions. However, how does that jive with this example, where there is one unique solution after comparing coefficients?
 
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Mr Davis 97 said:
I have the linear equation ##a + (2a - b) \sqrt{2} = 4 + \sqrt{2}##. I commonly hear that for linear equations, if we have two variables and one equation, then the system is undertermined and there are infinitely many solutions. However, how does that jive with this example, where there is one unique solution after comparing coefficients?
I may be missing something here, but what "unique solution" are you referring to? This is a linear expression so it should have an infinite amount of solutions, as there are no discontinuites in its graph.
 
Mr Davis 97 said:
I have the linear equation ##a + (2a - b) \sqrt{2} = 4 + \sqrt{2}##. I commonly hear that for linear equations, if we have two variables and one equation, then the system is undertermined and there are infinitely many solutions. However, how does that jive with this example, where there is one unique solution after comparing coefficients?
It seems to me that you are restricting your solution to the integers. Solving your equation with respect to a gives a=\frac{(b+1)\sqrt{2} + 4}{1+2\sqrt{2}}, which gives one value of a for each value of b you insert in the formula.
 
Mr Davis 97 said:
I have the linear equation ##a + (2a - b) \sqrt{2} = 4 + \sqrt{2}##. I commonly hear that for linear equations, if we have two variables and one equation, then the system is undertermined and there are infinitely many solutions. However, how does that jive with this example, where there is one unique solution after comparing coefficients?
It depends on what type of equation you're working with.
If there are no conditions on a and b, then there are infinitely many solutions, one of which is ##a=\sqrt 2, b = 0##.
If a and b must be integers, then the coefficient of ##\sqrt 2## must be the same on both sides of the equation, which yields the unique solution a = 4 and b = 7.
 
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Mr Davis 97 said:
if we have two variables and one equation, then the system is undertermined and there are infinitely many solutions.

Consider the single equation in two variables: ##a + b = a + b + 1##
 
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