# Is there an elementary “proof” that the first degree equation has only one solution?

I was asked from a student why the first degree equation has only one solution (if it has a solution)

Let's consider the equation $2x+5-3x=-4x+14$ for example.
How can I explain to a 13 year old student that the above equation does not have 2 or more solutions?

From my experience I believe that explaining "everything" to a student is not the best way to teach him/her mathematics, but I don't want at any case to give the impression that there are rules we use which cannot be explained.

• Such an equation could have infinitely many solutions. – David Mitra Dec 6 '14 at 10:42
• @DavidMitra How can such an equation have infinitely many solutions? It seems clear to me that the answer can only be $x=3$ unless you mean one with multiple variables like $x=y$. – Dom Dec 6 '14 at 10:53
• @Dom I meant, in general, a linear first degree equation can have infinitely many solutions. – David Mitra Dec 6 '14 at 10:57
• @Dom Take this example: $x=x$. It's clearly a first degree equation, and yet it has an infinite number of solutions! (Also relevant: $x=x+1$ is a first degree equation with no solutions.) – Akiva Weinberger Dec 8 '14 at 0:42

Firstly, try to explain the meaning of "iff". And then say that if we use basic properties like adding or multiplying both parts with nonzero number we get exactly the same equation. Thus our new equation has exactly the same roots. So at the end you will have something like x=a and you can say that every b=/=a doesn't satisfy the equations so there can't be other solutions.

This is potentially a good opportunity to illustrate why it is useful to have rules for arithmetic - that addition is commutative and associative and has an identity (zero) and inverses. Similarly for multiplication. And that the distributive law holds.

Whether these are taken as definitions of how numbers work, or are derived from other properties thought to be more fundamental is another discussion.

So let's begin with your equation:

$$2x+5-3x=-4x+14$$

First we use the commutative law for addition to rearrange the two sides so that all the terms in $x$ come before all the constant terms i.e. $$2x-3x+5=-4x+14$$

Now note that $4x$ is the additive inverse of $-4x$, so add this to both sides:

$$4x+2x-3x+5=4x-4x+14=14$$

Where we use the associative law to group $4x$ and $-4x$ together. Next we add $-5$ to both sides so that $$4x+2x-3x+5-5=4x+2x-3x=14-5=9$$

Now with $$4x+2x-3x=9$$ we start to invoke laws which involve multiplication. First we use the distributive law on the right-hand side $$(4+2-3)x=3x=9$$ Then we multiply both sides by $\frac 13$ to get $x=3$.

The key thing is that each of these laws is reversible, so the argument will go both ways.

In general this method can be used to put the equation in the form $ax=b$ - then either $a=0$, when $x$ can be anything, or we can multiply by $\frac 1a$ to get $x=\frac ba$.

It is the systematic use of the basic laws of arithmetic which renders this (the outline of) a proof.

You could tell them that first degree polynomials correspond to straight lines (and if someone asks why, then introduce the thing to them), so solving a first degree equation really means finding the $x$ coordinate of the point of intersection between two straight lines, which is obviously unique.