Certainty of being able to solve linear equation

In summary: This equation is not very helpful in determining a unique solution. In fact, any values of a and b that satisfy the equation are solutions. Therefore, the system is undertermined and has infinitely many solutions.In summary, the conversation discusses the concept of undertermined systems and infinitely many solutions in linear equations. It is commonly understood that a system with two variables and one equation is undertermined and has infinitely many solutions. However, this may not always be the case as seen in the example provided, where there is one unique solution after comparing coefficients. This is because the solution is restricted to a specific set of values, such as integers. In conclusion, the number of solutions in a linear equation depends on the type of equation and
  • #1
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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  • #2
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.
 
  • #3
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 [itex]a=\frac{(b+1)\sqrt{2} + 4}{1+2\sqrt{2}} [/itex], which gives one value of a for each value of b you insert in the formula.
 
  • #4
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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  • #5
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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