Certainty of being able to solve linear equation

[Mr Davis 97]

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?

 

[Comeback City]

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.

 

[Svein]

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.

 

[Mark44]

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.

 

[Stephen Tashi]

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: $a + b = a + b + 1$