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Solving algebra

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I'm using a program that teaches math, and it says to find x, you start from the right and work your way back. The example was: 2x + 5 = 7 7 - 5 = 2 2 / 2 = 1 x = 1 and I understand that, but how does this one work: 4 - x = -4 Can someone explain this case to me step by step? Cheers. :)

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for your first example i assume you dropped the x early? i.e.

2x + 5 = 7
7 - 5 = 2x
2 = 2x
2/2 = 2x/2
1 = x

it's just that it was a bit unclear, normally you keep your algabraic terms in the equation at all times.

in any equation you can add, subtract, divide or multiply both sides by anything, but you must do it to both sides. with that in mind...


4 - x = -4 lets add 4 to each side

4-x + 4 = (-4) + 4

8-x = 0 the two fours on the right cancel to give 0

now add x to each side

8 - x + x = (+)x
8 = x

I don't want to confuse you, but i think "start from the right and work your way back." is a confusing approach. It's better to try to get the x term on it's own, then you should be able to solve.

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I find it strange how your program solves that equation. I prefer to see it another way: you have your equation, on which you can perform operations, which lead to equivalent equations (i.e. an equation which yields the same solutions).

If you have
a = b
then you can add the same number to both sides (in this case, c):
a = b ===> a + c = b + c
Adding a negative number is the same as subtracting, so this means you can subtract the same value from both sides as well:
a = b ===> a - c = b - c

A simple example on this would be
x - 5 = 9 ==> (x - 5) + 5 = 9 + 5 ==> x = 14

The same can be done with multiplication (and multiplying by 1/x is the same as dividing, so you can also do it with division), but beware, multiplying (or dividing) by 0 is not allowed.
x/7 = 2 ==> (x/7) * 7 = 2 * 7 ==> x = 14


There are several ways you can solve your equation:
4 - x = -4 ==> (4 - x) - 4 = -4 - 4 ==> -x = -8 ==> -x * -1 = -8 * -1 ==> x = 8
4 - x = -4 ==> (4 - x) + x = -4 + x ==> 4 = -4 + x ==> 4 + 4 = (-4 + x) + 4 ==> 8 = x

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I'm closing the thread. The original poster got what he wanted, a step-by-step solution.

The original post, and both replies, were in violation of forum policy. The original post presented something right out of a textbook, very easily could be schoolwork, and asked flat out for a step-by-step solution rather than advise or hints. And both replies provided the requested step-by-step solution.

I know its kind of boring, and even unthought-of to read the forum FAQ, but in fact there is a very clear policy on questions that are indistinguishable from schoolwork---as these are. Please go back and read the policy, and try to follow it!

Forum FAQ

Thank you!

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