Solve the equation 2x + 2λx + μ = 0 for x, treating λ and μ as parameters.

Step-by-step solution

1. 1

   Collect the $x$ terms and factor. Both $2x$ and $2\lambda x$ carry a single power of $x$, so they combine:

   $2x+2\lambda x=2x(1+\lambda)$

   and the equation becomes

   $2x(1+\lambda)+\mu=0$

   This form is the whole point: the coefficient of $x$ is now a single visible quantity, $2(1+\lambda)$, and everything depends on whether it is zero.

2. 2

   Solve the generic case $\lambda\neq -1$. Then $2(1+\lambda)\neq 0$ and you may divide:

   $2x(1+\lambda)=-\mu\quad\Longrightarrow\quad x=\frac{-\mu}{2(1+\lambda)}$

   There is exactly one solution. For example $\lambda=3,\ \mu=5$ gives $x=-\frac{5}{8}$, and indeed $2(-0.625)+6(-0.625)+5=0$.

3. 3

   Handle the degenerate case $\lambda=-1$. Now $1+\lambda=0$, the $x$ term disappears entirely and the equation reduces to the constant statement

   $\mu=0$

   Nothing can be divided by zero here, so you must read the two sub-cases off directly.

4. 4

   Split the degenerate case in two. If $\lambda=-1$ and $\mu=0$, the equation is $0=0$: it holds for every real $x$, so the solution set is all of $\mathbb{R}$ — an identity, not an equation with one root. If $\lambda=-1$ and $\mu\neq 0$, the equation says $\mu=0$ with $\mu\neq 0$: a contradiction, so the solution set is empty.

5. 5

   Summarise the trichotomy. A linear equation $Ax+B=0$ always splits this way, and here $A=2(1+\lambda)$, $B=\mu$:

   $\lambda\neq-1:\quad x=\frac{-\mu}{2(1+\lambda)}\ \text{(unique)}$

   $\lambda=-1,\ \mu=0:\quad x\in\mathbb{R}\ \text{(infinitely many)}$

   $\lambda=-1,\ \mu\neq0:\quad \text{no solution}$

   Skipping the case split is the standard mistake: dividing by $1+\lambda$ without checking it is nonzero silently throws away the last two possibilities.