Unique solution when a≠0

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.