Linear equations

This is what lets us call the equation "linear": one unknown, one clean answer, never two, never none, as long as a is not zero.

Existence. Starting from $ax+b=0$, add $-b$ to both sides: this does not change the balance, and gives $ax=-b$. Since $a\neq0$, we may divide both sides by $a$: this too preserves the balance, and gives $x=-\dfrac{b}{a}$. Substituting this value back, $a\left(-\dfrac{b}{a}\right)+b=-b+b=0$, so it genuinely satisfies the equation — a solution exists.

Uniqueness. Suppose $x_1$ and $x_2$ are both solutions, so $ax_1+b=0$ and $ax_2+b=0$. Subtracting the two equations eliminates $b$: $a x_1 - a x_2 = 0$, i.e. $a(x_1-x_2)=0$. Because $a\neq0$, the only way a product of two numbers is zero is if the other factor is zero, so $x_1-x_2=0$, i.e. $x_1=x_2$. Hence no two different numbers can both solve the equation: the solution found above is the only one.

This shows the label "linear equation" secretly depends on the coefficient of x being nonzero — drop that, and the whole notion of "one unique answer" collapses into either every answer or no answer.

Substitute $a=0$ directly into $ax+b=0$: the term $ax$ becomes $0\cdot x=0$ for every real number $x$, since any number times zero is zero. So the equation literally becomes $b=0$, a statement about $b$ alone that no longer mentions $x$ at all.

Now there are exactly two possibilities for the fixed number $b$. If $b=0$, the leftover statement "$0=0$" is true regardless of which $x$ we substituted — so every real number $x$ satisfies the original equation, giving infinitely many solutions. If instead $b\neq0$, the leftover statement "$b=0$" is simply false — no value of $x$ can make a false numerical statement true, so the equation has no solution whatsoever, no matter what $x$ we try.