 # How can you move a fraction to the other side and have the same answer?

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Hey, I have a question about math.

If you take :

- 8 = 3/19 n (imagine the 3/19 as a fraction.)

you need to find N. To do this you would flip the 3/19 to get 19/3 and put it of the other side of the equation. Getting:

19/3 * -8/1 = n

Getting in the end the answer:

-152/3 = n

How does this give the answer?

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The moving-a-fraction-to-the-other-side-and-flipping-it is shorthand for the following steps (based on the fact that if you multiply both sides by the same number then the equality is still true):

1. multiply both sides of the equation by 19 - this cancels with the /19 on the RHS

2. divide both sides by 3 - this cancels with the 3 on the on the RHS

You are left with the LHS multiple by 19/3 and n on the RHS.

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Look: the principle is that when you have A = B then you can perform the same operation on both sides and the statement is still true.

By that I mean you can multiply both sides by 19/3

So -8.19/3 = 3/19 n . 19/3

I did that because that makes the numbers on the RHS cancel out, giving -152/3 = n. Easy

Edit - cross-posting with Strange, as usual

Edited by DrKrettin

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Edit - cross-posting with Strange, as usual ##### Share on other sites

The moving-a-fraction-to-the-other-side-and-flipping-it is shorthand for the following steps (based on the fact that if you multiply both sides by the same number then the equality is still true):

1. multiply both sides of the equation by 19 - this cancels with the /19 on the RHS

2. divide both sides by 3 - this cancels with the 3 on the on the RHS

You are left with the LHS multiple by 19/3 and n on the RHS.

Thanks.

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Hey, I have a question about math.

If you take :

- 8 = 3/19 n (imagine the 3/19 as a fraction.)

you need to find N. To do this you would flip the 3/19 to get 19/3 and put it of the other side of the equation. Getting:

19/3 * -8/1 = n

Getting in the end the answer:

-152/3 = n

How does this give the answer?

That's a property of linear equations.

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It is a linear equation, of the form "An= B" with A= 3/19 and B= -8. It is not a linear function so cannot be graphed- though you could think of it a point on the graph y= (3/19)x at y= -8.

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It is a linear equation, of the form "An= B" with A= 3/19 and B= -8. It is not a linear function so cannot be graphed- though you could think of it a point on the graph y= (3/19)x at y= -8.

Didn't think it worked like that.

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Here is the solution-

$-8=\dfrac{3}{19}n$

Now, multiply both sides by $19$

$-8\times 19=\not19 \times\dfrac{3}{\not19}n$

$-152=3n$

Now, divide both sides by $3$

$\dfrac{-152}{3}=\dfrac{3n}{3}$

$\dfrac{-152}{3}=n$

Edited by deesuwalka

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Back in my dark past, when I was teaching basic algebra, I always discouraged students from saying "move a number from one side of the equation to another".

With something like "ax= b" many students would come up with x= - b/a. Then when I pointed out that the answer should be b/a, they would complain "but I thought that when you moved a number from one side to the other you had to change the sign!".

Instead I tried to teach them to think in terms of "reversing" whatever is done to x. In ax= b, x is multiplied by a. To reverse that do the opposite: divide by 3. And, of course, whatever you do to one side you must do to the other: ax/a= x= b/a.

If the equation were x+ a= b, then, since a is added to x, we do the opposite: subtract a: x= a- b.

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Back in my dark past, when I was teaching basic algebra, I always discouraged students from saying "move a number from one side of the equation to another".

With something like "ax= b" many students would come up with x= - b/a. Then when I pointed out that the answer should be b/a, they would complain "but I thought that when you moved a number from one side to the other you had to change the sign!".

That would be from the students not actually thinking through the process and simply trying to get the answer. They must also understand how it works if they are to understand how to get the answer. Perhaps it gets them through, but it won't be useful in the long run.

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