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#1 2005-12-08 13:53:10


Linear Equations

I need help with linear equations.

1. Do linear equations follow the order of operations?

For example, x/3 + 6=12

2. When solving this, why do you have to subtract 6 first? (You are not following order of operations).
3. When you do the reciprocal first on the problem above, you get the wrong answer?
4. What is the difference between subtracting 6 first or doing the reciprocal first?


Just say you have a problem like


5. In order to solve, you can distrbute or you can do the reciprocal of the fraction 1/4 and then multiply it by both sides. When you are doing this equation: x/3 +6=12, when you do the reciprocal on both sides, I get the wrong answer why?

BTW, I accuratly know how to solve these, its just I want to understand the methods properly.


#2 2005-12-08 14:37:50

Registered: 2005-08-04
Posts: 394

Re: Linear Equations

It's because, when solving things algebraically, whatever you do to the equation, you must do to both sides--every part of both sides. For example:

x/3 + 6 = 12
x/3 + 6 - 6 = 12 - 6
x/3 = 6
3(x/3) = 3(6)
x = 18


x/3 + 6 = 12
3(x/3 + 6) = 3(12)
x + 18 - 18 = 36 - 18
x = 18

It really doesn't matter in what order you do things, but personally, I like the first method better. smile

El que pega primero pega dos veces.


#3 2005-12-08 14:39:22

Registered: 2005-11-28
Posts: 97

Re: Linear Equations

I'm not certain that I understand your question...but here goes...

The equation   x/3 + 6 = 12   is not multiplying the entire left side of the equation by 1/3....the only thing that is effected by that 1/3 is the x term...then, if you multiplied that entire side by 3, you would be making incorrect adjustments to the constant term (6)...

so if you wanted to use that method (multiplying both sides by the reciprocal) you would need to make the entire left side of the equation into just one fraction ---> (x+18)/3 = 12

I'm not sure that my explanation made sense...hope that helped

The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
                                                             -Bertrand Russell


#4 2005-12-09 16:41:03


Re: Linear Equations

Thanks guys! Ye, what ever you do to one side, you got to do to the other. Thanks again.

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