These types of questions really are easier than they look. Basically, when they say "solve for x" or "find the value of x," they mean to get the x by itself. That is, get the 'x' in the equation on one side of the equation and get everything else on the other side of the equation. The 'x' must be completely by itself. You can't have '-x' or '3x' or '3+x.'
So the first thing that needs to be done is to rewrite the equation with all the numbers with variables attached to them on one side of the equal sign and all the numbers without any variables attached (no 'x' attached) on the other side of the equal sign. To do this, minus 4 from both sides. Your problem should look like this:
6x+4-4=16+2x-4
When you work this out, you get this equation:
6x=12+2x
Now we need to move the '2x' from the right side of the equal sign to the left side. To do this, you simply minus 2x from both sides. Your equation should now look like this:
6x-2x=12+2x-2x
When you do 6x-2x and 2x-2x, you get this equation:
4x=12
Now we need to detach the 4 from the x. Do this by dividing both sides by 4. We can't minus 4 from both sides since the 4 is attached to the x rather than just sitting next to it. 4x really means 4 times x, not 4 plus x. Therefore, the 4 can't be removed by minusing. That would be like solving (2 times 2=x) by minusing 2 from both sides. So instead of getting 4=x, you'd get 2 times 2-2=x-2. Which is far different, right? Going back to the originaly question, after dividing both sides by 4, you'd get this equation:
(4x)/4=12/4
After simplifying this, you get your answer of:
x=3.
Let's do the next question. 7x-4=11+2x. Once again, we want to get the 'x' by itself on the left side of the equal sign. The first step is to get all the numbers with 'x' attached to it on the left side of the equal sign and all the numbers with no X's attached to it on the other side of the equal sign. There are a few steps to doing this. First, let's move the -4 to the right side. Unlike in the last example, we can't minus -4 from both sides. This would result in the left side looking like 7x-8 rather than 7x. Therefore, we have to add 4 to both sides. This way, the -4 turns into 0 (-4+4=0). If you do this, your equation should now look like this:
7x-4+4=11+2x+4
When simplified, the equation looks like this:
7x=15+2x
We still have the 2x on the right side. We want it on the left side. How do we move it over to the left side? Minus 2x from both sides. Therefore, we'll have this equation:
7x-2x=15+2x-2x
When simplified, we get this:
5x=15
Well look at what we have here, all the X's are on the left side and the numbers without X's on the right. Now all we need to do is get rid of the 5. Once again, we can't add or subtract since the 5 is attached to the x rather than sitting next to it. So how do we get rid of it? Like in the last problem, we divide both sides by 5:
(5x)/5=15/5
Which, when simplified, gets you this answer:
x=3
Woopie! We solved the problems! Respond to my answer if you still don't get it and we'll do some more problems like this.
2007-03-06 07:59:39
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answer #1
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answered by ender 3
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If you always keep in mind that these are equations, meaning each side will be equal to the other, then you should understand that whatever 'action' done to one side also must be done to the other.
6x + 4 = 16 + 2x
Let's get all the "x" values on one side of the equation, and all numerical values on the other side. So how can this be done?
On the left side, we can subtract the 4 to leave 6x by itself. But then we also must subtract 4 from the other side.
6x + 4 - 4 = 16 + 2x - 4. Let's simplify these to:
6x = 12 + 2x You understand that 4-4 is zero and 16 - 4 = 12
Now let's subract 2x from each side to have 12 by itself.
6x-2x = 12 + 2x - 2x
4x = 12 We can now divide each side by 4 to get x by itself.
4x/4 = 12/4
x = 3 Lets check our answer.
6 x 3 + 4 = 16 + 2x3
18 + 4 = 16 + 6
22 = 22
Picture an equation with each side in a pair of scales. You cannot solve the equation without using the other side. Each side will give us clues.
Look at the next one. 7x - 4 = 11 + 2x . Lets get all the x's on one side and the numerical values on the other for the quickest solution.
Subtract 2x from each side, which will give us:
5x - 4 = 11
Add 4 to each side.
5x = 15
Divide by 5 each side.
x = 3 Check work; 21 - 4 = 11 + 6, 17 = 17
2007-03-06 15:42:49
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answer #2
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answered by ? 6
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You do it like this:
problem: 6x+4= 16 +2x ... subtract 2x from 2x then subtract 6x from 2x. then it will be: 4x+4=16... now you subtract 4 from four and you do 16 from 4. now the equation is: 4x= 16
divide 4x by 4 and then divide 16 by four and your answer will be 4. x=4 and you follow the same steps for the second problem.
2007-03-06 15:36:34
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answer #3
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answered by ? 2
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6x + 4 = 16 + 2x
OK, first you have to isolate x.
You can subtract 4 from both sides which would give you:
6x + 4 - 4 = 16 + 2x -4
Which becomes:
6x = 12 + 2x
Now subtract 2x from both sides:
6x - 2x = 12 + 2x - 2x
Which becomes:
4x = 12
Now, to get x all alone, you will have to divide 4x by 4:
4x/4 = 12/4
x = 3
So, 7x - 4 = 11 + 2x
7x - 4 + 4 = 11 + 2x + 4
7x = 15 + 2x
7x - 2x = 15 + 2x - 2x
5x = 15
x = 3
2007-03-06 15:26:35
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answer #4
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answered by Anonymous
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6x + 4=16 + 2x
subtract 4 from each side
6x = 12 + 2x
then subtract 2x from each side
4x = 12
solve for x
4(3)=12
do the same for the next equation
2007-03-06 15:28:45
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answer #5
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answered by Steph 2
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6x+4=16+2x
-4 -4
6x=12+2x
-2x -2x
4x=12
x=3
7x-4=11+2x
+4 +4
7x=15+2x
-2x -2x
5x=15
x=3
Both variables are equal to 3
2007-03-06 15:26:11
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answer #6
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answered by Breaker's Lax Rules!!! 2
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First of all, in each problem, try to get the X's on one side of the equation and the integers on the other side.
They should come out like this:
4x=12
5x=15
Then you solve like you would any other equation
x=3
x=5
It's really pretty easy once you get it.
2007-03-06 15:19:55
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answer #7
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answered by Flappy 3
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move the x's onto one side and the the numbers onto the other side. then divide to get x by its self.
like move the 2x over by subtracting and then move the 4 over by subtracting again. then divide
2007-03-06 15:24:15
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answer #8
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answered by sdalsania 3
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