solving the domain.

• December 2nd 2010, 09:19 AM
bdubb
solving the domain.
Solve x = 8x - 42 over the domain {0,2,4,6}
• December 2nd 2010, 09:21 AM
emakarov
Well, I am not sure what "solving over the domain" is, but moving x to the right and 42 to the left gives 7x = 42, from where x can be found.
• December 2nd 2010, 10:02 AM
bdubb
Quote:

Originally Posted by emakarov
Well, I am not sure what "solving over the domain" is, but moving x to the right and 42 to the left gives 7x = 42, from where x can be found.

I still don't understand.:(
• December 2nd 2010, 10:06 AM
Plato
Quote:

Originally Posted by bdubb
I still don't understand.:(

What is there not to understand?
You simply must tell in detail what difficulty you are having with problem.
Can you solve $7x=42~?$
• December 2nd 2010, 10:13 AM
emakarov
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.
• December 2nd 2010, 10:40 AM
Bacterius
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 $\{0, 2, 4, 6\}$. Typically, you would first solve the equation completely then discard any solution that doesn't fit within the subset, so let's try that :

$x = 8x - 42$

$42 = 8x - x$

$42 = 7x$

$x = \frac{42}{7} = 6$

$6$ is the solution and it is within the domain, thus $6$ 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 $7$. The number $7$ is not in the domain $\{0, 2, 4, 6\}$, therefore you need to discard it.

Does it make sense ?
• December 2nd 2010, 10:43 AM
bdubb
Quote:

Originally Posted by Bacterius
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 $\{0, 2, 4, 6\}$. Typically, you would first solve the equation completely then discard any solution that doesn't fit within the subset, so let's try that :

$x = 8x - 42$
Yes now i understand! thanks!!
$42 = 8x - x$

$42 = 7x$

$x = \frac{42}{7} = 6$

$6$ is the solution and it is within the domain, thus $6$ 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 $7$. The number $7$ is not in the domain $\{0, 2, 4, 6\}$, therefore you need to discard it.

Does it make sense ?

Yes, so it would be 6
• December 2nd 2010, 10:47 AM
Bacterius
But it wouldn't be $7$, or $8$, or $-4$, 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 $\{0, 2, 4, 6\}$ are solutions to the actual question. Any other number needs to be discarded.

(In logic that would be written $S_{\textrm{question}} = \{0, 2, 4, 6\} \cap S_{\textrm{equation}}$ ...)