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When solving for unknowns why does the symbols sometimes change from $a + $ to $a -$ or vice versa or stays the same?

For example (using solutions from text): $2x+6 = x+16\quad$ and the solution is $2x-x = 16-6\quad \Rightarrow \quad x=10$.

Or

$8x-12 = -3x+21\quad $Solution: $8x+3x = 21+12\quad \Rightarrow \quad 11x = 33\quad \Rightarrow \quad x=3$

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4 Answers

Here are some of the steps left out.

$$2x+6-x-6=x+16-x-6$$ $$(2x-x)+(6-6)=(x-x)+16-6$$

Does this help?

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No not really, where are you getting all the 6's from there only one in the equation? and in my text book it says to move the unknown variables to the left and move the numbers to the right. I just don't understand how it starts off adding then switches to subtraction? –  Dan Nov 22 '12 at 18:52
    
@Dan The variables are moved by adding or subtracting to both sides of the equation to eliminate it from one side. So to eliminate $x$ from the right, subtract both sides by $x$. Similarly for the 6. Then combine like terms. –  Mike Nov 22 '12 at 19:02
    
so by switching the x and the 6 the positive becomes a negative? please bear with me as this kinda of math is kinda of new to me. –  Dan Nov 22 '12 at 19:15
    
@Dan You eliminate $x$ on one side by subtracting both sides by $x$ (or adding $-x$ if you prefer) since $x-x=x+(-x)=0$ –  Mike Nov 22 '12 at 23:18
    
Well, what about. $2x+6−x−6=x+16−x−6\to2*x+1*6+(-1)*x+(-1)*6=1*x+1*16+(-1)*x+(-1)*6$ Since we have everything in addition now, we can rearrange them at our will. That is how the minuses and pluses work. (Or I think they do) –  Yellow Skies Nov 23 '12 at 0:26
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For example: $2x+6 = x+16$, and the solution is $2x-x = 16-6\Rightarrow x=10$.

Here, let's start by adding $-x$ to both sides of the equation $2x +6 = x+16$: $$-x + 2x+6 = -x + x + 16$$ $$\leftrightarrow x+6 = 16.$$

Next: Adding $-6\,$ to each side of the equation gives us: $$x + 6 + (-6) = 16 + -6$$ $$\leftrightarrow x + 0 = 16 - 6 $$ $$\leftrightarrow x = 10$$

Solving for $x$ given $8x-12 = -3x+21$:

Add $3x$ to each side of the equation, AND add $12$ to both sides of the equation. This gives: $$3x + 8x -12 + 12 = 3x + -3x + 21 + 12$$ $$\leftrightarrow 11x + 0 = 0 + 21 + 12$$ $$\leftrightarrow 11x =33$$ $$\leftrightarrow x = 3.$$

Note that in the last step, we divided both sides of the equation by $11$ (which is the same as multiplying both sides of the equation by $\frac{1}{11}$).

The objective when solving for one unknown, say $x$, is to get all multiples of $x$ on the left hand side, and all constants on the right hand side. Then simplify. As long as you are adding (or subtracting) the same value from each side of the equation, and as long as you multiply (or divide) both sides of an equation by the same value, the equality remains unchanged.

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These equations are from the text book and the solution is already answered in the text book, but they dont clarify how they change from addition to subtraction? I'm upgrading my math so some of this is new to me so bear with me. –  Dan Nov 22 '12 at 19:06
    
so how does the values stay the same or change? –  Dan Nov 22 '12 at 19:26
    
What remains unchanged is the equality of each side. What I'm describing is a way to simplify the equation to solve for $x$: which means representing the equation in the form $x = \text{blah}$. –  amWhy Nov 22 '12 at 19:31
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wouldn't X=10 and not 16 since 16-6=10? –  Dan Nov 22 '12 at 19:35
    
Yes Dan, sorry for the typo! –  amWhy Nov 22 '12 at 22:59
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The basic principle of high school and middle school algebra is the use of inverse operations.

What you're seeing here is repeated use of the fact that $+$ has the inverse operation $-$ and $\times$ has the inverse operation $\div$.

More technically speaking, we see that these equations are solvable because there are inverse elements of each element in the equation. We can undo any operations that are present in the equation, which allows us to find the solution for $x$ or whatever variable is present.

I find that quadratic equations really explain just how important these fundamental observations are. We take it for granted in middle school, high school, and sometimes college that for a polynomial $ax^2+bx+c=0$ there are two solutions: $$x=\frac{-b+\sqrt{b^2-4ac}}{2a} \text{ and } x=\frac{-b-\sqrt{b^2-4ac}}{2a}.$$

However, think about this: What if there was no such thing as $\frac{1}{2a}$? As absurd as this question may seem to you now, this kind of thinking is very important in higher math. It is the existence of the inverse element $\frac{1}{2a}$ (among other elements) that allows us to solve these quadratic equations. This same situation occurs in linear equations and other equations.


At the risk of sounding stupid or uninformed (I'm not entirely confident with abstract algebra), I'd like to explain just why the above questions and others are so relevant. The quadratic formula doesn't always work. Yes, in high school, it always works. But why? Because we're considering the equation $ax^2+bx+c=0$ with $a,b,c,x$ in the field $(\mathbb{C},+,\cdot)$. You may not know what a field is right now, but I can give you an idea of why it's an interesting concept and how it relates to the essence of your question.

The field $(\mathbb{C},+,\cdot)$ is the collection of all elements which are composed of usage of addition and multiplication on complex numbers. (Division and subtraction are technically not relevant: We can add any negative number to perform subtraction and likewise multiply any fraction to divide.) What's important here is this: The characteristics of the field you're working in determines what equations are solvable! That is the beauty of abstract algebra (well, one of the many) to me. It is no longer simply, "We know that $x$ is blah blah." It's that we know there is always going to be a solution to particular sets of equations on the basis of the characteristics of our field.

I'm rambling a bit here, but I hope this was enlightening!

P.S. All of you on Math.SE that are way more enlightened than me, please tell me if I've butchered any of the technicalities. I apologize profusely if that is the case, and I'll correct it as swiftly as possible. Thank you.

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Nice work John. You shouldn't, ever, feel inferior to others. Math.SE is not for Research Mathematics—it's for everybody. You can also find people as dumb as me. :) –  Parth Kohli Nov 23 '12 at 15:12
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Thanks for the compliment. You're right: Math.SE is for everyone. :) –  000 Nov 23 '12 at 18:52
    
Why don't you come to OpenStudy any more? –  Parth Kohli Nov 24 '12 at 5:35
    
@ColorfulTrauma, between my many activities in and outside of academia, I find it very difficult. It is very hard to find questions I enjoy on that site as well. I've sent you a message telling you the more precise details. :) –  000 Nov 24 '12 at 7:20
    
That's why I have left that site as well. ;) –  Parth Kohli Nov 24 '12 at 7:22
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First, you must understand that if we have a statement like $n = m$, then $n$ equals $m$. That means that $n + 1 $ is STRICTLY equal to $m$. That also means that $2m = 2n$, ${n \over 2} = {m \over 2}$ and whatnot.

What you always do in linear equations is, in fancy term, “isolate” the variable. What is isolation? Isolating a variable is to perform certain actions to both sides (important) in order to have a variable on one side and a constant on another, if possible. I have an example later, but before...

Your basic rule is that you have to do the same thing on both sides. You can do anything, but for it to lead you to an answer, you have to follow a suggested rule: “$\rm AS-MD$”

  • Addition and subtraction in any order.
  • Division and multiplication in any order.

Note: you apply this rule to the variable side—the side which has your variable required to be solved for.

Let's have an example now. $$2x + 1 = 5$$An effective way is to subtract one from both sides according to the rule.$$\begin{align}2x + 1 - 1& = 5 - 1 \\ 2x&=4\end{align}$$Now that we have nothing to subtract or add, let's try the multiply/divide part. Yes, we can divide $2$ from both sides. $${2x \over 2} = {4 \over 2} \quad \Rightarrow \quad x = 2$$


Now, that you get the basic idea, let's try to solve the same equation through the way you suggested. $$\begin{align}2x + 1 &= 5 \\ \\ 2x & = 5 - 1 \\ \\ 2x & = 4 \\ \\ x &= {4 \over 2} \\ \\ x & = 2 \end{align}$$ Now, I cannot prove rigorously, but I can tell you that the sign changing seems something fancy but is based on the idea of doing the same things on both sides. The intution: $$a + m = b \quad \Rightarrow \quad a = b-m $$ Which is derived by, $$\begin{align} a + m - m & = b - m \\ \\ a + \underbrace{( m - m)}_{\text{always zero}}& = b - m \\ \\ a &= b - m\end{align}$$

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