Solve x = 8x - 42 over the domain {0,2,4,6}
If you have an equation a = b, you can add c to both sides to get a + c = b + c. This is just like if you have weighing scales that are balanced, you can put equal weight on both sides while preserving the balance. Similarly, you can subtract c from both sides of the equation a = b to get a - c = b - c.
So we start with x = 8x - 42. Subtract x from both sides: x - x = 8x - 42 - x, or 0 = 7x - 42. Next, add 42 to both sides: 0 + 42 = 7x - 42 + 42, or 42 = 7x. Here, in fact, we divide both sides by 7 to get 42 / 7 = 7x / 7, or 6 = x.
To understand this, you need to know how to rearrange and simplify algebraic equations, such as adding like terms.
Edit: You can also substitute each of the numbers 0, 2, 4, 6 for x in the equation x = 8x - 42 and see which of them turns it into a true equality.
I think he doesn't understand the "domain" part. Basically, it means you need to solve the equation within the subset given to you which here is a discrete set of integer values, namely $\displaystyle \{0, 2, 4, 6\}$. Typically, you would first solve the equation completely then discard any solution that doesn't fit within the subset, so let's try that :
$\displaystyle x = 8x - 42$
$\displaystyle 42 = 8x - x$
$\displaystyle 42 = 7x$
$\displaystyle x = \frac{42}{7} = 6$
$\displaystyle 6$ is the solution and it is within the domain, thus $\displaystyle 6$ is the solution to the question.
But that question is quite retarded in the sense that it doesn't properly show the concept of restricted domain ... because imagine the equation above had another integer solution (not true but it's an example) which was for instance $\displaystyle 7$. The number $\displaystyle 7$ is not in the domain $\displaystyle \{0, 2, 4, 6\}$, therefore you need to discard it.
Does it make sense ?
But it wouldn't be $\displaystyle 7$, or $\displaystyle 8$, or $\displaystyle -4$, if these solutions came up, that's what's important to understand in this question. Only solutions to the equation that are within the set $\displaystyle \{0, 2, 4, 6\}$ are solutions to the actual question. Any other number needs to be discarded.
(In logic that would be written $\displaystyle S_{\textrm{question}} = \{0, 2, 4, 6\} \cap S_{\textrm{equation}}$ ...)