I want to know why in a system of linear equations I'm allowed to sum or subtract the equations. I can't get the intuition of why I can do that to solve for the equations.
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Consider
\begin{align*} 5 + 6 &= 11 \quad \quad(1) \\ 7 + 12 &= 19 \quad \quad(2) \\ (1) - (2)\implies (5 - 7) + (6 - 12) &= 11 - 19 \\ -2 - 6 & = -8 \quad \text{(true)} \end{align*}
You see that for equations not containing unknowns, the equality of both sides is not violated upon subtraction / addition.
You are allowed to do the same thing to equations containing unknowns because they adhere to the same rules of arithmetic and get replaced by real numbers after the system has been solved.
hickslebummbumm
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If $a=b$ and $c=d$, then $a-c=b-d$. This is obviously true as you are doing the same computation on both sides of the equal sign. For the same reason, $3a-2b=3c-2d$.
Now if your equalities are complete equations, $ax+by+cz=d$ and $ex+fy+gz=h$, there is no reason you can't do the same
$$(ax+by+cz)-(ex+fy+gz)=d-h,$$which can be rewritten
$$(a-e)x+(b-f)y+(c-g)z=d-h.$$
Also
$$(3a-2e)x+(3b-2f)y+(3c-2g)z=3d-2h.$$
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The intuition is the following:
If you have an equation, say, $$2x + 3 = 4$$ you know you are allowed to add the same thing to both sides, or subtract the same thing from both sides, or multiply both sides by the same thing, or divide both sides by the same thing. For example, an equivalent equation where you add $5$ to both sides is: $$2x + 3 + 5 = 4 + 5 $$
So, you are comfortable with doing that. Well, when we have another equation, say, $$x + 3 = 19$$ what we are saying is that the quantity $x + 3$ is the same as the quantity $19$ (since we wrote $x + 3 = 19$). Well, if they are the same, then adding $x + 3$ to the left hand side of $2x + 3 = 4$, and adding $19$ to the right hand side, should be allowed because we are allowed to add the same thing to both sides of an equation. It just so happens that our second equation tells us that $x + 3$ and $19$ are the same thing.
That's why we can say $$2x + 3 + (x + 3) = 4 + 19$$ -- because all we did was add the same thing to both sides of the equation.
So, that's the intuition for adding one equation to another, and why it is allowed.
layman
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I only have one last question, why if I solve the equation for x it happens that the x is the x where both intercept – kprincipe Oct 21 '14 at 00:06
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@kprinciple Well, take a look at the two equations in my example. The first one was $2x + 3 = 4$ and the second one was $x + 3 = 19$. If I want to find $x$ that makes both of these equations true, then I am going to pretend that there is such an $x$. Then I will assume that with this $x$, we have $x + 3$ is the same thing as $19$. If this is the case, then when I add $x + 3$ on the left hand side and $19$ on the right hand side of the equation $2x + 3 = 4$, and solve for $x$, the $x$ that I get will solve the equation $2x + 3 = 4$, but because we used $x + 3$ as if it was equal... – layman Oct 21 '14 at 00:33
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to $19$, then our solution will also satisfy $x + 3 = 19$. I know this might be hard to understand at first. But we aren't just solving $2x + 3 = 4$. You could easily solve that equation on your own if you wanted to without involving another equation. But as soon as we start toying with $x + 3 = 19$, too, then we get the answer that satisfies both equations. Does it make any more sense now? – layman Oct 21 '14 at 00:35
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Yeah it makes more sense now. – kprincipe Oct 21 '14 at 01:04
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Actually sorry to bother but there's something I still don't get and is that I thought that if i sum the equation 1 to the 2 I could use the resulting equation to evaluate the same values as with the equation 2 because I was adding the same to both sides of equation but I can't get the same results. Hope I am explaining well – kprincipe Oct 21 '14 at 10:47
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@kprincipe The example I gave in my answer is a bad one because there is only one variable in the equations. This means there is no solution that satisfies both of the equations at the same time in the equations I gave. You know this because you solved it the way we said, and your solution doesn't satisfy both equations. I should have used an example with two variables. Here is an example: Suppose we want to solve the system: $\begin{cases} 2x + 3y = 1 & \text{} \ 2x + 4y = 2 & \text{} \ \end{cases}$. You can add or subtract one equation from the other to cancel out the $x$'s... – layman Oct 21 '14 at 12:04
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and then solve for $y$, and plug that value back into one of the two equations to solve for $x$. The solution you get, which will be two numbers, $x$ and $y$, should satisfy both equations (you can check by plugging the $x$ and $y$ into each equation above and making sure you get what is on the left hand side). – layman Oct 21 '14 at 12:06
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I think i didnt explained well in this example you have the system of equations if i take equation 2 and add 5 to both sides it still the same equation if i let y alone and evaluate x i would get the same results from equation 2 and equation 2 plus 5. I thought that this would be the same if i add equation 1 to both sides of equation 2 the new equation i obtain i can use it to evaluate x values and get the same results as using equation 2. That is my doubt because loke you said in the beginning adding equation 1 to both sides of equation 2 is like adding 5 to both sides . – kprincipe Oct 21 '14 at 19:42
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@kprincipe I don't understand your question. Can you maybe provide an example of what you are talking about? – layman Oct 22 '14 at 00:28
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For example if i have 2x + 3y =1 and 2x + 4y = 2. If i add 5 to both sides of equation 1 it would be the same equation so if i let y alone and evaluate any x for the equation 1 or the equation 1 after adding 5 to both sides i get the same results. My question is if equation 2 is 2x + 4 and i add ot to the equation 1 can i do the same. I mean with the resulting equation of adding the 2 equations if i let y alone why i dont get the same y's as with equation 1 cause all i do was add the same on both sides of equations like with the 5 – kprincipe Oct 22 '14 at 00:53
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@kprincipe I'm still having trouble understanding what you are saying. I will say this: for the first equation, we can find many solutions that make $2x + 3y = 1$. For example, if $x = -1, y = 1$, this will satisfy the equation. If $x = 4, y = -\frac{7}{3}$, this also works. We can find many more. But they don't all work for the second equation $2x + 4y = 2$. So, to find out which ones work for both, we subtract the first equation from the second and solve for $y$. The action of subtracting one equation from another is what makes the solution work for both equations. – layman Oct 22 '14 at 00:58
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Ok, If I have (1) $2x + 3y = 1$ and (2) $2x + 4y = 2$ is the same as (1) $y = {1-2x\over 3}$ and (2) $y= {2-3x\over 4}$. I thought that if I added equation (1) to equation (2) the result is (3) $4x + 7y = 3$ If I solve this equation for $y$ will get (3) $y = {3-4x\over 7}$ my question was why if I evaluates $x$ in (3) will not get the same values as equation (2) cause I added equation (1), the same to both sides as adding 5 to both sides of equation (2) and solving for $y$, the 5 will cancel. I thought the same would be but instead adding 5, adding another equation. – kprincipe Oct 22 '14 at 11:26
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@kprincipe You made a mistake. (1) $2x + 3y = 1$ and (2) $2x + 4y = 2$ is the same as (1) $y = \frac{1 - 2x}{3}$ and (2) $y = \frac{2 - 2x}{4} = \frac{2(1 - x)}{4} = \frac{1 - x}{2}$. – layman Oct 22 '14 at 11:59
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think of it like this. Suppose you have a=b and c=d where a,b,c,d can be equations based on different variables. So we could say that if a=b then a-b=0 ,since a and b are the same thing. Same for d-c=0. But now we have a-b=0=c-d, thus a-b=d-c. Now we can add b and c on both sides, thus we get a+c=d+b. did that answer your question?
Liealgebrabach
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