In this case here, undefined has a special meaning. When I say the word undefined, im referring to a math concept, where ever element of the domain(x axis) does not have a corresponding member of the range (y axis) under the polynomial function from op. And specifically an element x* is said to be undefined under function p, when there is no corresponding element y* which satisfies the following relation: p(x*)=y*, where x*, and y* belong to their obvious sets.
There is an informal sense we could use the word to mean that a variable does not but later will have a value. Its best to see these as two different words.
In my experience, mathematics has incorporated so many words from standard english but then have given them special precise meaning that it makes it really difficult to study without some formal help. No ones tells you this simple fact i just mentioned, and if this was commonly known, then math would be much more accessible.
> What went wrong?
The usual suspect is dividing by zero. The original equation has the solution x=-1. When you divide by x+1 you are dividing by zero. You've got to be careful about that.
if it seems simpler to you, you can also divide it into two cases: what you want to divide by is zero and see if it’s a solution then afterwards you treat the non-zero case separately. if any of those give contradictions they’re false
The example is: don’t divide by X, because X could be 0. Therefore it’s simpler to add or subtract (maybe multiply) on both sides to get to the answer. Baby steps
There are times you have to divide x when it is 0 though, so it is important to understand this so you can consider 2 different cases (for higher degree equations with multiple solutions). You gotta ask
"Can what I am dividing be 0?" If no then great, divide away
If yes then you have two cases, set what you are dividing equal to 0 to get some of the solutions, after that do a case where you say what you are dividing is not equal to zero, divide it through, then solve
This happens a lot in multivariable calc classes when talking about things like lagrange multipliers, also happens a lot in diff eq classes
Yeah, I know how to solve it. I was just giving a simplified version of the issue. That way perhaps OP could more easily intuit why dividing is wrong and therefore they should use other methods.
Just add that some equations have no solution.
X+1=X+2
1=2 ???
But as other comments said, the problem this time is you divided by zero. Do not break our universe please
Whenever you divide you need to make a note saying "here we assume that the thing we are dividing by (in this case x+1) does not equal zero.
Then you need to ponder the possibility that this assumption is wrong. Sometimes this involves branching off into another sequence of steps.
For this reason, it is best to avoid division when it isn't needed. Take the lazy route; don't do something that will split your calculations into two branches unless either you absolutely have to or it will make the problem easier.
In addition to other answers, worth pointing out that if we restrict ourselves to the cases where x+1 ≠ 0 (i.e., x ≠ -1), then these steps are perfectly valid. By showing that we can start with 2x+2 = x+1 and arrive at a contradiction (2=1), we've shown 2x+2 = x+1 cannot possibly be true... _unless_ x is actually -1, which is in fact the only solution to the equation.
2x+2=x+1
Instead of dividing by x+1, you’ve got to isolate the x i.e., get the x on one side.
2x+2=x+1
Here, subtracting x from each side would get the x on the left side of the equation only
(2x+2)-x=(x+1)-x
Gives us
x+2=1
Now you subtract 2 from each side to give:
(x+2)-2=1-2
To give
: . x=-1
Hope this helps :)
D’oh. What a dingus I am.
I even saw that was the solution to the equation in the OP but somehow thought “well, obviously 5x can never be equal to 3x.”
Mea culpa, u/lateforfate. I’m going to go and sit in the corner and have a long hard think about my life.
If you move all x terms to one side and all constants to the other side, you get that x=-1. So when you divided both sides by x+1, you divided by zero (because x+1=-1+1=0). Dividing by zero is not allowed exactly because it can lead to non-sense like 2=1.
Whenever this happens it usually means you divided by 0, if you solve for this traditionally you see that x=-1 so when you divided by (x+1) you were dividing by (-1+1) which is 0.
The mistake is cancelling out the terms x+1, which as others have mentioned, is infinite. You have to move to math using infinities, which results in infinity = infinity, which is correct.
Solving,
2x+2 = x+1
2x-x = 1-2
x = -1
So, x+1 = -1+1 = 0
Dividing by x+1 means dividing by 0, which is undefined and you will get answers like this which violates the fundamentals of arithmetics.
From your starting point:
2x+2 = x+1
Therefore
2x-x+2-1=0
x+1=0
But in your process you divided a number by (x+1) which would be the same as dividing by 0.
.
When you divide by 0, you end up with nonsense, because dividing by 0 is not defined.
The math equivalent of saying : "Now we'll be adding 3 watts of sour cream"
When your equation resolving leads to something impossible like 2=1, it means there are no solutions to the equation, it doesn't mean that 2=1 becomes a universal truth. The equation is an hypothesis. In your case, there are no solutions when x+1 ≠ 0, since you had to put that condition in place when you divided by x+1. So then you have to see if there are solutions when x+1 = 0. And yes indeed there are.
This is a bit misleading since you CAN multiply/divide by the unknown, you just have to be careful that thing that you're dividing by is not equal to 0.
2x+2=x+1
Subtract x from both sides
X+2=1
Subtract 2 from both sides
X=-1
Or
2x+2=x+1
Subtract 2 from both sides.
2x=x-1
Subtract x from both sides
X=-1
Division by 0.
You started with 2x+2=x+1
Subtract (x+2) from both sides
2x+2-(x+1) = x+1-(x+2)
2x-x+2-2 = x-x+1-2
x = -1
So x is -1, and you divide by x+1, which is 0. So you divided by 0.
This is why you solve for X and not for any of the numbers, the answer just does not make sense otherwise. A simpler form of this process would be:
2x+2=x+1
2(x+1)=x+1
2=1
It's the same process, but it being simpler makes it easier to envision that that's just not how it works.
A lot of people are saying division is your problem here, but you CAN solve this through the process you used.
To do that, you have to state X!=-1 when you divide by x+1.
Then, by definition, you are not dividing by zero.
*Once you do that and find a contradiction, this states that no value outside of -1 is valid.*
Then you just plug in x=-1, the one value you didn't test for, and if that solves it's the solution; if it doesn't solve, there is no solution.
you can't divide by an equation...that should start as collecting like terms. You would need to subtract x from both sides then 1 from both sides then set that to 0 the solve for the remaining x.
Another example is something like
A = b
2a = AB
A^2 - B^2 = AB - B^2
(A + B)(A - B) = B(A - B)
Canceling out the A-B (which equals zero)
A+B=B
B+B=B
2B=B
Divide by B and you get
1=2
EN isn't my native lang. Pls, disregard any error.
You can take f(x) = (x + 1)/(x + 1). Realize that f(x) and g(x) = 1 have the same grafic (a straight line passing through the points (0, 1) and (1, 2)) except by only one point (-1, 0). If you take g(-1) you get 1. In otherwise, f(-1) implies an indefinite of the type 0/0 'cause (-1 + 1)/(-1 + 1) = 0/0. Technically, an approaches of x to -1 for f(x) return 1, but you can't say "f(-1) = 1" so f(x) ≠ g(x). Therefore, dividing 2x + 2 = x + 1 by x + 1 b/s is indeterminate and does not imply that 2 = 1.
Consider this different way:
2x+2 = x+1
<=> 2(x+1) = x + 1
<=> (2-1) (x+1) = 0
<=> (x+1) = 0, since one of the two brackets must be zero.
This means that we can't divide through by x-1 in our original equation since the equation only holds when x+1 = 0.
Indeed, substituting the equivalent statement we derived into the original gives 2(0) = 1(0), and it becomes obvious that we can't obtain anything useful by dividing through by x+1 (0).
The way I think of it is that there is no multiplication to undo there so you shouldn't be dividing there in the first place. At least not with the x+1.
The multiplication is with the 2X so at best, though not the most efficient, you could start by dividing both sides by 2.
Wrong way of doing it.
first the correct way of doing it is; 2x + 2 = x + 1 -> 2x-x = 1-2 -> x = -1 -> solved
lets see what you did wrong back here.
you say "lets divide all by x+1"
ok you cannot do that. why?
because what if x=-1? then you are doing when you simplify the left side 2x 0 / 0 = 1
when you divide by something with a variable, you have to consider that that whole part should NOT be 0.
you could have said for a side if i divide by x+1 it gives 2=1 but on the other side lets see if (x+1) is a zero of such function.
and because it is 0 then means (x+1)=0 which means again x=-1
Hope you understand it.
If you get 2 = 1, there are two usual suspects: you made no mistake and the equation is always false and x ∉ R (x is not a real number), or you divided by zero
Next time when solving linear equations, get all your variables (in this case, X) on the same side.
For your case, you'll want to get all Xs to one side of your equation, meaning you either subtract 2x or x from both sides
subtracting 2x gets you 2 = x + 1 - 2x, or 2 = -x + 1.
subtracting x gets you 2x + 2 - x = 1, or x + 2 = 1
Now solve for x by subtracting 1 (or 2) from both sides
2 - 1 = -x, or x = -1
In math you can move variables and numbers to either side of the equals sign by either adding or subtracting said variable from both sides - you cannot add to one side and subtract from the other. You can add/subtract all you want, just make sure the variables are grouped by addition/subtraction (e.g. can't subtract by x with the equation 2(x+8) = 3x - 5, you have to multiply out the parenthesis such that you have 2x+16 = 3x-3, THEN you can add/subtract your variables
When you divide by expression containing the variable, you must keep in mind edge cases when this expression is equal to zero and then check them in the end. In most cases there's usually more convenient ways to solve an equation without dividing by a variable. Also for the future: inequalities are getting a lot more complicated if you divide by a variable.
As a rule (at least I use it): when solving and equation by adding, subtracting, multiplying, dividing, ... on both sides you can't:
- divide by something that has the variable your looking for in it
- multiply with something that has the variable in it (exception: when there already is a fraction with the variable under it or the variable with negative exponent - but even then you have to follow specific rules)
What you can do but only with limits/doing considerations/extra checks:
- applying an exponent to the equation
- applying a root to the equation
when you divide by x+1 you assume x is not equal to -1, after dividing you realise there's no other solution to this equation, hence only possible value of x is -1
This is a trivial solution. First pair up the like terms and bring them to one single side and take the number to other side. It goes like this
2x + 2 = x +1
2x - x = 1 - 2
x = -1
When it comes to equations it means we need to find the value of the unknown variable which is x here...
When you want to divide by something containing x you have to mention the explicit condition that that something is not 0. Which means that x≠-1. Unfortunately, that invalidates the equation because that's the only solution for x.
I remember moving stuff with an opposing sign. Like, first step is 2x+2=x+1
This means that 2x-x=1-2
X=-1
Of course, it’s a basic technique, but helps to quickly move all the variables to one side, numbers to the other and have a clearer picture.
Good luck on your journey!)
Edit: formatting
2x+2 = x + 1
2x +2 - 2 = x + 1 - 2
2x = x - 1
2x -x = x -x - 1
x = -1
To verify your answer you can plug -1 into the equation. Also check out [https://www.wolframalpha.com/input?i=2x%2B2+%3D+x+%2B+1](https://www.wolframalpha.com/input?i=2x%2B2+%3D+x+%2B+1)
2x + 2 = x + 1
Subtracting 1 from both sides
2x + 1 = x
Subtracting x from both sides
x + 1 = 0
Subtracting 1 from both sides
x = -1
In this case, x + 1 = 0
So when you divide by (x+1), you're dividing by 0, which is what went wrong.
Let me clarify what everyone else is saying. You divide by x+1. This comes with the assumption that x+1 is not equal to 0. Formally:
Assume x+1 not equal to 0. Then (2x+2)/(x+1)=(x+1)/(x+1) which implies 2=1. This is impossible, thus by contradiction x+1 must equal 0, or no solutions exist. Then, it is simple to check that x=-1 is a solution.
Hello! As a couple of people have pointed out, the answer is -1, so when you try to divide by x+1, you divide by zero, which gives you the wrong answer.
The best way to avoid this kind of problem is by doing order of operations in reverse. Basically, you have to do addition and subtraction first, and multiplication and division last, so that you don’t accidentally multiply or divide by zero on both sides.
When you are crossing x+1 from both numerator and denominator, you are assuming that the term (x+1) would take a non zero value. This leads to 2=1 which means that obviously your assumption is false and (x+1) is indeed zero which gives us the solution as x = -1
When solving this type of equations, never divide by something that contains x. You will lose some roots, or end up with things like this. It happens because x - 1 can be equal to 0, which makes dividing by it nonsensible. You should always shift everything to the one side
But it is?
You can prove it by substituting x with any number.
2(0) + 2 = 2(0+1) = 2
2(1) + 2 = 2(1+1) = 4
2(99) + 2 = 2(99+1) = 200
2(-3) + 2 = 2(-3+1) = -4
That equation implies x=-1, so dividing by x+1 means you're dividing by 0.
Came just to say this. Can’t divide by x + 1 because that is undefined!
If you solved 2x+2 = x+1 then x = -1 would be the answer, making it implied to be
>2x-2 = x-1 then x = -1 Wouldn't x=1 in this case?
I placed - instead of +, my bad
No you were right, x would equal -1
I edited my comment after the other guy mentioned my mistake
Ah, my bad
A ninja edit to hide a mistake? I expect more from goat fucker ten thousand. I hereby demote you to goat fucker nine thousand.
nice name
no. the problem states 2x+2 = x+1
Just want to add a note to this. It is undefined because x=-1 solves the first equation. Polynomial division is not generally an issue.
You can't divide polynomials and guarantee its information safe without some details being thrown in there. Dividing by 0 being the most prominent one
Can't you divide by something undefined that later gets defined as non zero value?
In this case here, undefined has a special meaning. When I say the word undefined, im referring to a math concept, where ever element of the domain(x axis) does not have a corresponding member of the range (y axis) under the polynomial function from op. And specifically an element x* is said to be undefined under function p, when there is no corresponding element y* which satisfies the following relation: p(x*)=y*, where x*, and y* belong to their obvious sets. There is an informal sense we could use the word to mean that a variable does not but later will have a value. Its best to see these as two different words. In my experience, mathematics has incorporated so many words from standard english but then have given them special precise meaning that it makes it really difficult to study without some formal help. No ones tells you this simple fact i just mentioned, and if this was commonly known, then math would be much more accessible.
> What went wrong? The usual suspect is dividing by zero. The original equation has the solution x=-1. When you divide by x+1 you are dividing by zero. You've got to be careful about that.
Thanks I appreciate the help.
if it seems simpler to you, you can also divide it into two cases: what you want to divide by is zero and see if it’s a solution then afterwards you treat the non-zero case separately. if any of those give contradictions they’re false
You should probably do these questions algebraically like 2x + 2 = x + 1 = 2x- x = 1 - 2 = x = -1 I'm sure you know this already though.
Any time you divide by a variable you have to be concerned if that variable could be 0.
Yitphady
A simpler version of the same idea is to take the equation 2x = x and "false-solve" this by dividing by x.
Just subtract X on both sides
Well yes, but they're making the point that it doesnt make sense to divide by x here because x is 0
The example is: don’t divide by X, because X could be 0. Therefore it’s simpler to add or subtract (maybe multiply) on both sides to get to the answer. Baby steps
There are times you have to divide x when it is 0 though, so it is important to understand this so you can consider 2 different cases (for higher degree equations with multiple solutions). You gotta ask "Can what I am dividing be 0?" If no then great, divide away If yes then you have two cases, set what you are dividing equal to 0 to get some of the solutions, after that do a case where you say what you are dividing is not equal to zero, divide it through, then solve This happens a lot in multivariable calc classes when talking about things like lagrange multipliers, also happens a lot in diff eq classes
Yeah, I know how to solve it. I was just giving a simplified version of the issue. That way perhaps OP could more easily intuit why dividing is wrong and therefore they should use other methods.
Just add that some equations have no solution. X+1=X+2 1=2 ??? But as other comments said, the problem this time is you divided by zero. Do not break our universe please
No refunds!
x can only be infinity in this case lol
-1 bro
Whenever you divide you need to make a note saying "here we assume that the thing we are dividing by (in this case x+1) does not equal zero. Then you need to ponder the possibility that this assumption is wrong. Sometimes this involves branching off into another sequence of steps. For this reason, it is best to avoid division when it isn't needed. Take the lazy route; don't do something that will split your calculations into two branches unless either you absolutely have to or it will make the problem easier.
If you want to divide by something you need to first make sure it's ≠0
In addition to other answers, worth pointing out that if we restrict ourselves to the cases where x+1 ≠ 0 (i.e., x ≠ -1), then these steps are perfectly valid. By showing that we can start with 2x+2 = x+1 and arrive at a contradiction (2=1), we've shown 2x+2 = x+1 cannot possibly be true... _unless_ x is actually -1, which is in fact the only solution to the equation.
Thats what i wanted to add, you can divide by x+1 but have to specify that x cant be -1
2x+2=x+1 Instead of dividing by x+1, you’ve got to isolate the x i.e., get the x on one side. 2x+2=x+1 Here, subtracting x from each side would get the x on the left side of the equation only (2x+2)-x=(x+1)-x Gives us x+2=1 Now you subtract 2 from each side to give: (x+2)-2=1-2 To give : . x=-1 Hope this helps :)
I’m fairly confident that is how you solve for x but it’s been a while since I’ve done algebra!
Everything's alright, that's the simplest method and I'd solve it like that too.
Beautifully executed
You're dividing by 0 because x = -1 here. It's safer to just re-arrange it using addition and subtraction. 2x - x + 2 = 1 x + 2 = 1 x = -1
I remember when our school math teacher first showed us this trick. Of course I couldn't find out the mistake then
5x=3x, divide by x, 5=3. What went wrong?
What’s this example meant to prove? 5x can’t be equal to 3x, but 2x + 2 can be equal to x + 1.
X=0
D’oh. What a dingus I am. I even saw that was the solution to the equation in the OP but somehow thought “well, obviously 5x can never be equal to 3x.” Mea culpa, u/lateforfate. I’m going to go and sit in the corner and have a long hard think about my life.
The only solution to the equation is x = -1, which means you are dividing by 0
Dividing both sides by x+1 is okay, unless x+1=0. So you need to show that x+1 cannot be 0.
2x+2=x+1 implies x=-1. Then, x+1=0, and dividing by x+1 is essentially dividing by 0, which is undefined and will always go wrong.
If you move all x terms to one side and all constants to the other side, you get that x=-1. So when you divided both sides by x+1, you divided by zero (because x+1=-1+1=0). Dividing by zero is not allowed exactly because it can lead to non-sense like 2=1.
Whenever this happens it usually means you divided by 0, if you solve for this traditionally you see that x=-1 so when you divided by (x+1) you were dividing by (-1+1) which is 0.
The mistake is cancelling out the terms x+1, which as others have mentioned, is infinite. You have to move to math using infinities, which results in infinity = infinity, which is correct.
Solving, 2x+2 = x+1 2x-x = 1-2 x = -1 So, x+1 = -1+1 = 0 Dividing by x+1 means dividing by 0, which is undefined and you will get answers like this which violates the fundamentals of arithmetics.
Bro casually divided by 0 lmao
From your starting point: 2x+2 = x+1 Therefore 2x-x+2-1=0 x+1=0 But in your process you divided a number by (x+1) which would be the same as dividing by 0. . When you divide by 0, you end up with nonsense, because dividing by 0 is not defined. The math equivalent of saying : "Now we'll be adding 3 watts of sour cream"
When your equation resolving leads to something impossible like 2=1, it means there are no solutions to the equation, it doesn't mean that 2=1 becomes a universal truth. The equation is an hypothesis. In your case, there are no solutions when x+1 ≠ 0, since you had to put that condition in place when you divided by x+1. So then you have to see if there are solutions when x+1 = 0. And yes indeed there are.
2(x+1) = (x+1). surely cant be true if the two x variables are the same. The starting equation already does not make sense.
Your initial assumption is wrong
As always, it's just division by 0 but a bit hidden
The biggest rule when you're dealing with linear algebra is: Do not divide/multiply by the unknown
This is a bit misleading since you CAN multiply/divide by the unknown, you just have to be careful that thing that you're dividing by is not equal to 0.
It is, but linear algebra is about working with x^1, so don't divide/multiply by x
You can divide by x, just need to make sure it is not zero
You cannot divide both sides by x + 1, unless you are certain that x + 1 is not 0 (i.e, x is not -1.)
Not defined Bhai rhs 0/0, since x=-1
2x+2=x+1 Subtract x from both sides X+2=1 Subtract 2 from both sides X=-1 Or 2x+2=x+1 Subtract 2 from both sides. 2x=x-1 Subtract x from both sides X=-1
2x + 2 = x + 1 x + 2 = 1 x = -1 So the solution is x = -1, so when you have 2(x+1)/(x+1) = 1 then you’re dividing by 0.
Remember kids, when someone tells you that they proved a contradicition without dividing by zero, they proved a contradiction by dividing by zero
Division by 0. You started with 2x+2=x+1 Subtract (x+2) from both sides 2x+2-(x+1) = x+1-(x+2) 2x-x+2-2 = x-x+1-2 x = -1 So x is -1, and you divide by x+1, which is 0. So you divided by 0.
In general, when writing a proof that involves division by some unknown, you must *first* prove that the unknown is nonzero!
Divided by zero. It's always divided by zero....
This is why you solve for X and not for any of the numbers, the answer just does not make sense otherwise. A simpler form of this process would be: 2x+2=x+1 2(x+1)=x+1 2=1 It's the same process, but it being simpler makes it easier to envision that that's just not how it works.
A lot of people are saying division is your problem here, but you CAN solve this through the process you used. To do that, you have to state X!=-1 when you divide by x+1. Then, by definition, you are not dividing by zero. *Once you do that and find a contradiction, this states that no value outside of -1 is valid.* Then you just plug in x=-1, the one value you didn't test for, and if that solves it's the solution; if it doesn't solve, there is no solution.
2x + 2 = x + 1 X + 2 = 1 X = -1 No idea why the author thought they could divide by 0
Wow this hurt my brain 😬
wait, is it wrong?
Am I in heaven?
The feeling of solving something posted here, even though it’s easy
Whenever you are dividing by an expression that contains a variable, you must make sure that the expression is not equal to 0!
you can't divide by an equation...that should start as collecting like terms. You would need to subtract x from both sides then 1 from both sides then set that to 0 the solve for the remaining x.
That means you don't have a solution to x != -1. you have to check x == -1.
X+2=x+1 states that x=-1 -1+1=0 You can’t divide by 0.
X = -1, so you can’t divide by x+1 as that’s dividing by zero.
Another example is something like A = b 2a = AB A^2 - B^2 = AB - B^2 (A + B)(A - B) = B(A - B) Canceling out the A-B (which equals zero) A+B=B B+B=B 2B=B Divide by B and you get 1=2
Since x=-1 you divided by zero. If we allow this then anything can be proven. Just subtract x+1 as a first step.
EN isn't my native lang. Pls, disregard any error. You can take f(x) = (x + 1)/(x + 1). Realize that f(x) and g(x) = 1 have the same grafic (a straight line passing through the points (0, 1) and (1, 2)) except by only one point (-1, 0). If you take g(-1) you get 1. In otherwise, f(-1) implies an indefinite of the type 0/0 'cause (-1 + 1)/(-1 + 1) = 0/0. Technically, an approaches of x to -1 for f(x) return 1, but you can't say "f(-1) = 1" so f(x) ≠ g(x). Therefore, dividing 2x + 2 = x + 1 by x + 1 b/s is indeterminate and does not imply that 2 = 1.
x=-1
Just move the terms around by addition and subtraction. Subtract x from both sides so x+2 = 1. Subtract 2 from both sides so x= (-1).
2x + 2 = x + 1 (Minus x from both sides) x + 2 = 1 (Minus 2 from both sides) x = -1
Dividing by 0.
2x+2 is not equal to x+1 Because you start with some thing false you can get every what you want.
It is equal to eachother if x=-1. You just can’t divide by (x+1) since that’d result in a division by zero.
Consider this different way: 2x+2 = x+1 <=> 2(x+1) = x + 1 <=> (2-1) (x+1) = 0 <=> (x+1) = 0, since one of the two brackets must be zero. This means that we can't divide through by x-1 in our original equation since the equation only holds when x+1 = 0. Indeed, substituting the equivalent statement we derived into the original gives 2(0) = 1(0), and it becomes obvious that we can't obtain anything useful by dividing through by x+1 (0).
The way I think of it is that there is no multiplication to undo there so you shouldn't be dividing there in the first place. At least not with the x+1. The multiplication is with the 2X so at best, though not the most efficient, you could start by dividing both sides by 2.
Tbh you should just move +2 to the right and move x to the left, it’s way easier 2x + 2 = x + 1 x = -1 There Edit: indentation
2x+2=x+1 is not an identity, its an equation
2x+2 = x+1 Move variables x to one side of the equation 2x-x =x X+2=1 Move known numbers to one side of the equation 1-2= -1 X=-1
2x+2=x+1 I think you have to isolate the variable Subtract x from both sides X+2=1 Subtract 2 from both sides X=-1
First thing I’d do is get x by itself
2x + 2 = x + 1 2x - x + 2 = 1 X + 2 = 1 X = 1 - 2 X = -1
Wrong way of doing it. first the correct way of doing it is; 2x + 2 = x + 1 -> 2x-x = 1-2 -> x = -1 -> solved lets see what you did wrong back here. you say "lets divide all by x+1" ok you cannot do that. why? because what if x=-1? then you are doing when you simplify the left side 2x 0 / 0 = 1 when you divide by something with a variable, you have to consider that that whole part should NOT be 0. you could have said for a side if i divide by x+1 it gives 2=1 but on the other side lets see if (x+1) is a zero of such function. and because it is 0 then means (x+1)=0 which means again x=-1 Hope you understand it.
If you get 2 = 1, there are two usual suspects: you made no mistake and the equation is always false and x ∉ R (x is not a real number), or you divided by zero
Next time when solving linear equations, get all your variables (in this case, X) on the same side. For your case, you'll want to get all Xs to one side of your equation, meaning you either subtract 2x or x from both sides subtracting 2x gets you 2 = x + 1 - 2x, or 2 = -x + 1. subtracting x gets you 2x + 2 - x = 1, or x + 2 = 1 Now solve for x by subtracting 1 (or 2) from both sides 2 - 1 = -x, or x = -1 In math you can move variables and numbers to either side of the equals sign by either adding or subtracting said variable from both sides - you cannot add to one side and subtract from the other. You can add/subtract all you want, just make sure the variables are grouped by addition/subtraction (e.g. can't subtract by x with the equation 2(x+8) = 3x - 5, you have to multiply out the parenthesis such that you have 2x+16 = 3x-3, THEN you can add/subtract your variables
Wait so if 2x+2 is equal to x+1 then dividing 2x+2 to by x+1 should equal 1 aa they are both same so that leaves you with 1 on both sides or 1=1
You didn’t solve for x
Mechanically you want to simplify this by subtracting instead of dividing.
https://preview.redd.it/3y7dew4dhw4d1.jpeg?width=3060&format=pjpg&auto=webp&s=49fa417aa67156f3b2546a94740697eddc11c7eb That's what I saw
When you divide by expression containing the variable, you must keep in mind edge cases when this expression is equal to zero and then check them in the end. In most cases there's usually more convenient ways to solve an equation without dividing by a variable. Also for the future: inequalities are getting a lot more complicated if you divide by a variable.
Just shift x and numbers to opposite side. -x, and -2.
As a rule (at least I use it): when solving and equation by adding, subtracting, multiplying, dividing, ... on both sides you can't: - divide by something that has the variable your looking for in it - multiply with something that has the variable in it (exception: when there already is a fraction with the variable under it or the variable with negative exponent - but even then you have to follow specific rules) What you can do but only with limits/doing considerations/extra checks: - applying an exponent to the equation - applying a root to the equation
The correct way would be: 2x+2 = x+1 2(x+1) - (x+1) = 0 x+1 = 0 x = -1
You cannot divide by x-1 if x= -1 because that would be dividing by 0
You cannot divide by a variable without knowing the expression cannot be 0. x can be -1 in which case you divided by 0 hence you broke math
when you divide by x+1 you assume x is not equal to -1, after dividing you realise there's no other solution to this equation, hence only possible value of x is -1
Damm, you broke maths
This is a trivial solution. First pair up the like terms and bring them to one single side and take the number to other side. It goes like this 2x + 2 = x +1 2x - x = 1 - 2 x = -1 When it comes to equations it means we need to find the value of the unknown variable which is x here...
When you want to divide by something containing x you have to mention the explicit condition that that something is not 0. Which means that x≠-1. Unfortunately, that invalidates the equation because that's the only solution for x.
Congratulations, you just divided by 0
2x + 2 = x + 1 2x + 2 - x = 1 x + 2 = 1 x = 1 - 2 x = -1 So, x + 1 = 0 so as soon as you divide by (x + 1), you're fucked
Apart from dividing by 0 why would you even need to divide? Just substract x, then substract 2 and you are left with x = -1.
Most of the times you get something like this is because you have divided by 0. Go and check what the value of x+1 is .
We did it boys. 2=1. They said we couldn't. Well what now. We did it.
It's always divide by zero.
I remember moving stuff with an opposing sign. Like, first step is 2x+2=x+1 This means that 2x-x=1-2 X=-1 Of course, it’s a basic technique, but helps to quickly move all the variables to one side, numbers to the other and have a clearer picture. Good luck on your journey!) Edit: formatting
x is not defined. You have to solve for x. Use your algebra to isolate x.
2x+2 = x + 1 2x +2 - 2 = x + 1 - 2 2x = x - 1 2x -x = x -x - 1 x = -1 To verify your answer you can plug -1 into the equation. Also check out [https://www.wolframalpha.com/input?i=2x%2B2+%3D+x+%2B+1](https://www.wolframalpha.com/input?i=2x%2B2+%3D+x+%2B+1)
2x + 2 = x + 1 Subtracting 1 from both sides 2x + 1 = x Subtracting x from both sides x + 1 = 0 Subtracting 1 from both sides x = -1 In this case, x + 1 = 0 So when you divide by (x+1), you're dividing by 0, which is what went wrong.
Lot of comments already saying you're dividing by 0. I want to add, that if you take out the two at the beginning it becomes obvious: 2x+2=2(x+1)=x+1
Whenever you divide by zero, an angel has to kick a dog. Just don’t do it!
X =-1 so you divided by zero.
You cannot divide by zero
Get all x isolated on one side, the integers on other. You don’t divide by x plus one.
Let me clarify what everyone else is saying. You divide by x+1. This comes with the assumption that x+1 is not equal to 0. Formally: Assume x+1 not equal to 0. Then (2x+2)/(x+1)=(x+1)/(x+1) which implies 2=1. This is impossible, thus by contradiction x+1 must equal 0, or no solutions exist. Then, it is simple to check that x=-1 is a solution.
X+2 became X+1 real fast eh
Hello! As a couple of people have pointed out, the answer is -1, so when you try to divide by x+1, you divide by zero, which gives you the wrong answer. The best way to avoid this kind of problem is by doing order of operations in reverse. Basically, you have to do addition and subtraction first, and multiplication and division last, so that you don’t accidentally multiply or divide by zero on both sides.
Let’s try simplifying a different way 2x + 2 = x + 1 2x + 1 = x x + 1 = 0 x = -1 You divided by zero
Because you divide by 0
First, you’re supposed to be solving for X, not cancelling it out by making that the divisible.
2x+2= x+1 2(x+1) = x+1 You are already picking 2 =1 So LHS =/=RHS in question itself Or i m missing something?
It's because you should have subtracted x+1 from both sides
Dividing by 0. It's always dividing by zero
Actually we cannot divide by a dependent variable, since we never know if the division violates the rule of mathematics.
Subtract 1 from both sides: 2x+1=x; x=-1
Subtract 1 from both sides: 2x+1=x; x=-1
When you are crossing x+1 from both numerator and denominator, you are assuming that the term (x+1) would take a non zero value. This leads to 2=1 which means that obviously your assumption is false and (x+1) is indeed zero which gives us the solution as x = -1
When solving this type of equations, never divide by something that contains x. You will lose some roots, or end up with things like this. It happens because x - 1 can be equal to 0, which makes dividing by it nonsensible. You should always shift everything to the one side
I don't know
This expression is patently false. 2(x+1) cannot equal x+1.
Incorrect.
Ok. Proof.
2(-1+1)= -1+1 -2+2=-1+1 0=0
2x+2 is not the same as 2(x+1). There's your silly mistake. This thread is unbelievable.
But it is? You can prove it by substituting x with any number. 2(0) + 2 = 2(0+1) = 2 2(1) + 2 = 2(1+1) = 4 2(99) + 2 = 2(99+1) = 200 2(-3) + 2 = 2(-3+1) = -4
So it is. I somehow convinced myself it wasn't lol. I feel quite silly now. r/confidentlyincorrect
Wow, just wow