Algebra and X's?

Question

I don't understand how you remove the X's in a sum...
19+8+13+x+13-x+28-x+8-x=77
89-2x = 77
-2x = 77-89
-2x = -12
x = 6

how did you get the 2x?

and this one:

10+x+15-x+13+x+10-x+12-x+28+x =93
how do you solve this.

thanks.

I 4got to add an extra X in the last equation...so it's :
10+x+15-x+13+x+10-x+12-x+28+x+x =93

Answer 1

-2x + 89 = 77

You get -2x because you have one positive x and 3 negative x's in the equation. So you just add them together 1x + -3x = -2x

Then move subtract 89 from both sides to get the x's by themself

-2x = -12

Now divide by -2 to get the x by itself

x = 6

That one was right

For the next one

88 + 3x - 3x = 93

Now combine the x terms, once you do that you have no x's left so this equation is false

88 = 93

So there is no solution for this one

I hope this helps

Answer 2

1st problem:
19 + 8 + 13 + x + 13 - x + 28 - x + 8 - x = 77
Restating the arrangement (group separately the x's from integers:
(x - x - x - x) + (19 + 8 + 13 + 13 + 28 + 8) = 77
There is 1 positive x and 3 negative x:
(+ x - 3x) + 89 = 77
Then it is understood why there is - 2x, that is x - 3x:
- 2x + 89 = 77
- 2x = 77 - 89
- 2x = - 12
x = - 12 / - 2
x = 6

I hope this clarifies things especially the - 2x.

2nd problem:
10 + x + 15 - x + 13 + x + 10 - x + 12 - x + 28 + x = 93
(x - x + x - x - x + x) + integers = 93
+ 3x - 3x + integers = 93
0x and we cannot anymore go on. There is no solution.

Answer 3

You have to combine like terms. There is an invisible one in front of all the x's. So in the last problem it's like this:

10 + 1x + 15 - 1x + 13 + 1x +10 - 1x + 12 - 1x + 28 + 1x = 93
put all the x terms together and all the constants together

10 + 15 + 13 + 10 + 12 + 28 = 78
1x - 1x + 1x - 1x - 1x + 1x = 0x (it all cancels)

So, you get 78 = 93 because the x's cancel and you get NO SOLUTION.

In the first example you have:
19 + 8 + 13 + 13 + 28 + 8 = 89
1x - 1x - 1x - 1x = -2x (you must keep the signs in front of the terms)

So, you have 89 - 2x = 77
Subtract 89 from both sides of the equation you get -2x = -12, then divide both sides of the equation by -2, so x = 6.

Answer 4

19+8+13 +(x) +13 -(x) +28 -(x) +8 -(x) =77
i have separated out the "x" parts of it
+1-1-1-1 = -2
-2x

19+8+13+13+28+8 = 89
89 -2x = 77
-2x = -12
-2x/-2 = -12/-2
x = 6


second equations has something wrong as all "x's" add to 0
leaving 88 = 93 which is not correct

Answer 5

X is a variable which represents a certain number which you are try to determine. In the first equation, you have to combine all the x's. In this case, you are adding one x and subtracting 3 x's (x - x - x - x), so you get -2x.

In the 2nd equation, combine the x's. You are adding 3 x's and subtracting 3 x's you you end up with no x's. But this will leave only 10+15+13+10+12+28 which is equal 88, not 93. So the original equation is not valid.

Answer 6

Add the numbers first.
89 + x - x - x - x = 77
89 + x - 3x = 77
89 - 2x = 77
89 = 77 + 2x
89 - 77 = 2x
12 = 2x
x = 6

Answer 7

Ignore the numbers in the first line.
Then you have x - x - x - x = x - 3x = -2x.
See, you owe me 1 dollar. I owe you 1, and then another, and then another. So I end up owing you 2 dollars.


Just group the xs together. For example, I have 1 cow, take a cow away, add a cow, take one away, take one away, and then add another one, I'm left with 0 cows. For your last one, no value of x will work because you're not left with any xs, but you are left with 88 = 93, which is never true.

Answer 8

All you are doing here is combining like terms.

(a nice rule to remember in algebra is that you can only add like terms that have the same exponent if they are a variable (ie. x + x = 2x but x^2 + x does not equal 2x.)


The problem with the 2nd question is that the x's cancel eachother out, it does not equal 93, therefore its unsolvable.

Answer 9

19+8+13+ ( x )+13- ( x )+28- ( x )+8- ( x ) =77

same as above, add teh numbers and add the x 's

PS: you guys.. thats a whole lot of fkin explainin for just 1-2= 1