# B Logical process of solving equations

1. Feb 1, 2017

### Mr Davis 97

I read that in solving an equation, the reault is only a possibility for a solution, and not necessarily a solution. For example, in solving x+3 = 5, x=2 is not a solution until you explicitly plug it back in to show that it actually. Why is the statement x=2 not enough to show that 2 is actually a solution, from a strictly logical point of view? This could apply to any equation-solving process.

2. Feb 1, 2017

### newjerseyrunner

Where did you read that and in what context? Could it be that your book used an example more like x^2 + 3 = 5? An equation like that simply doesn't have one solution.

3. Feb 1, 2017

### Orodruin

Staff Emeritus
This depends on what steps you take during your solution. If all steps are equivalencies rather than implications, you do not need to plug in your solution - unless your system of equations is overdetermined. In that case you need to check the equations you did not use.

An example of an implication step which is not an equivalence would be squaring the equation $x = 2$ to obtain $x^2 = 4$. You then introduce a false root $x = -2$.

Edit: That being said, it is of course never a bad thing to reinsert your result to check that your algebra was correct.

4. Feb 1, 2017

### Mr Davis 97

This is in the context of an abstract algebra course. My text says "Strictly speaking, we have not shown here that -3 is a solution, but rather that it is the only possibility for a solution. To show that -3 is a solution, one merely computes 5 + (-3)"

5. Feb 1, 2017

### Orodruin

Staff Emeritus
This still tells us nothing unless you specify the problem that was to be solved and the logical steps taken in doing so.

6. Feb 1, 2017

### Mr Davis 97

The problem was 5+x=2. And the logical steps are what you would expect (using inverse, identity, and associativity)

7. Feb 1, 2017

### Staff: Mentor

This is correct. But as you performed equivalence deductions, you already have proven the backward direction. In general, however, this might not be the case, so you will have to plug in the deduced statement to see, if it also satisfies the original equation, because a deduction only provides a necessary condition.
$$x+3=5 \Longleftrightarrow x=2$$
On the other hand, you could start with $x^3+y^3=z^3$ and find some necessary conditions on $x,y,z \in \mathbb{Z}$. But whatever you will find, it cannot satisfy the original equation.

Edit: Or more obvious: $2x= 4 \Longrightarrow 4x^2=16 \Longrightarrow x \in \{-2,2\}$ but $x=-2$ isn't a solution. (Oops, already been said.)

Last edited: Feb 1, 2017
8. Feb 1, 2017

### Mr Davis 97

So if you only use equivalence deductions (i.e. reversible operations) then the result is necessarily a solution because you could just argue in the reverse direction, while if the deductions are only one way (i.e. non-reversible), then you have to check solutions?

9. Feb 1, 2017

### Staff: Mentor

Yes. But it's always a good idea to check the solutions, especially if the transformations are manifold or complicated. "Reversible" isn't the exact term here, because you could reverse the squaring in the example above, but it generates an extra value which isn't a solution. So it has to be equivalent, not reversible.

Another famous pit is if you multiply an equation or greater-relation by a term, which isn't a known number at the time, e.g. something like $xy+z^2$ or similar, and you don't rule out the possibility that it might be zero or negative, then you get wrong results without knowing.

10. Feb 1, 2017

### FactChecker

Strictly speaking, you have proven in one direction but not in the reverse direction.
You have proven "If 5+x=2, then x=-3."
But to prove that "-3 is a solution of 5+x=2", you want to prove "If x=-3, then 5+x=2."
In this example, it seems like a silly distinction but in complicated examples it can be an important distinction.

11. Feb 2, 2017

### Stephen Tashi

Algebraic manipulations abbreviate a complicated process of reasoning. Solutions can be deduced by using "logical equivalences". Doing "reversible operations" in the sense of "doing the same operation to both sides" does not necessarily produce an equation that is logically equivalent to the original equation. Try using reversible operations to solve these equations:

1) $x + \frac{1}{x-2} = 2 + \frac{1}{x-2}$

2) $x \ln(x-2) = \ln(x-2)$

12. Feb 2, 2017

### pwsnafu

Consider $\sqrt{x^2+x+3} = x$.

13. Feb 2, 2017

### PeroK

We've been through this before with you!