Solving an equation of real numbers We have been given what seems to be a fairly easy task in calculus.
The answers in the book are not similar to mine, however.
$$
|7 - 3x| = -2
$$
I would assume the correct answer is: $$(0.6, 3)$$
The answer given is: "No solution"
The book has a reputation of having a rather large amount of incorrect answers, which is the reason why I am asking.
Thanks in advance for answer.
 A: Recall that $|x|$ is always larger than zero for any $x$. Essentially what the question is asking is
"For what $x$ is the absolute value of $x$ negative 2?"
Here we use $7-3x$ instead of $x$. The answer is that there is clearly no such value of $x$ that would yield a negative absolute value (because absolute values are never negative).
A: The answer " no solution " is correct. 
The reason is that $|7-3x|$ represents the absolute value of $(7-3x)$ which is greater than or equal to 0. 
Hence $|7-3x| \ge 0 \,\ \forall \,\ x$ i.e. $|7-3x| \not = -2$ for any real $x$.
A: The good thing about equations is that you can check your answers. You claim that $x=2$ is a solution. Well, let's plug it in: $|7-3\cdot 2| = |1| = 1 \neq 2$. So it turns out $x=2$ isn't a solution, and you can check that $x=0.6$ isn't either.
Now, I can't say much more without knowing how you arrived at your answers. But you can arrive at "no solution" immediately by noticing that no matter what the value of $x$ is, $|7-3x|$ can never be negative.
A: HINT: The absolute value of any real number can never be negative 
Hence, in general $$|x|\ge 0\ \ \ \forall\ \ \ x\in \mathbb R$$
A: I can see how you got that answer: you stripped away the absolute value signs, got the two equations $7-3x=-2$ and $-(7-3x)=-2$, and solved them. 
However, stripping off absolute value signs can introduce spurious solutions. 
It's just like solving radical equations: taking a power to get rid of the radical can introduce spurious solutions.
A: Hopefully this will make it obvious that there are no solutions, sketching modulus functions (by hand) to solve equations is helpful even for trivial cases such as this.
As seen below, just to be absolutely clear the $\color{blue}{\text{blue}}$ line is $x=-2$ and definitely doesn't intercept the $\color{red}{\text{red}}$ lines which is $y=|7 - 3x|$:

