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I was wondering why it is almost impossible to find a geometrical explanation of why adding two linear equations helps us to find a solution of a system of linear equations? Am I right that adding two linear equations will result in an equation of a line which will pass through a point where two linear equations intersect?

If it is right then I completely don't understand why such a crucial point in understanding how to solve systems of equations is never properly taught even at a university level.

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Your posts basically comes down to 2 questions:

1) Is the sum of 2 linear equations a line?

Answer: no not really. The result could be a plane, or a hyperplane, or the empty set.

Consider adding : $x+y = 1$ and $x + z = 1$, the result being $2x + y + z = 2$, that is a 3D plane.

Or consider adding $x = 1$ to $-x = 2$, the resulting graph being the empty set.

2) Does the intersection of two linear equations exist in the sum of the two linear equations?

Answer: if and only if the two linear equations have a non empty intersection; which only occurs if the variable coefficients are linearly independent or if the equations are equivalent.

For example: $x + y = 1$ and $2x + 2y = 3$ are not linearly independent (as far as the coefficients of x and y are concerned, as $[1, 1]\cdot 2 = [2,2])$. So the sum of this pair is $3x + 3y = 4$ and none of those points will satisfy either of the original equations.

As far as a crucial point that tends to be missing from university level "systems of equations", I'll just briefly mention that you should know that $x + y = z$ is a shorthand for $\{ [x,y,z] \in \mathbb R^3 \mid x + y = z \}$. This way of writing things can usually be avoided, but there may be times when you have to treat the sets of solutions as actual sets to understand certain lines of reasoning.

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Suppose we have two linear equations, $$y_1=m_1x+b_1$$ and $$y_2=m_2x+b_2.$$

If we set $y_1=y_2$,then $m_1x+b_1=m_2x+b_2$. We then solve for $x$ and substitute that value into either equation to get the corresponding $y$ value. The result is the single intersection point of these two straight lines. It will exist provided $m_1 \neq m_2$. I have just made an online interactive plot for you to play with. Sometimes this visual representation is what helps a concept like this to be truly understood.

Try simultaneously solving various equations like these while at the same time dragging the sliders to the appropriate values in this demonstration I have constructed. I think with this, clarity will eventually come to you.

An answer to the actual question:

If we simply add the equations side by side, we get a third equation that has slope equal to the sum of slopes, and $y$ intercept equal to the sum of $y$ intercepts. You can see this by plotting $y_1+y_2$ in the demonstration. This would be a third line that does not necessarily pass through the intersect of the two equations. In order to find the solution to a system of equations, that being the point at which both equations are simultaneously true, we need to set equals to equals, ie. $x$ equal to $x$ or $y$ equal to $y$. In this manner we find the solution to the system.

I have two more plots to show you plot 1 is the first, and plot 2 is the other. I think I understand precisely why you are asking this question now after reflecting on what you must be studying. It has to do with operations on systems of equalities, and how after multiplying one equality by a constant, then adding it to the other, you arrive at an equation in one variable, and thus arrive at a solution. You see that the sum intersects when a scalar multiple of one equality is added to the other. If you can manage to add a scalar multiple of one equation to the other in such a way that one of the variables vanishes, then you will have that third equation, it being the sum of two equations (not the originals), and that third equation will indeed intersect at the intersection of the first two equations. Best of clarity and fortune in your studies. :)

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  • $\begingroup$ This is the great demo, is it possible to some how ask it to show a line which will be a sum or difference of two already existing lines? $\endgroup$ – Rustam Issabekov May 4 '14 at 6:27
  • $\begingroup$ Absolutely! Just click below and make the next line say "y_1+y_2" and see what happens. I was just doing that myself hehe. You can do lots of exploring in this simple interface. Note that I made the default slopes additive inverses, so you will get a horizontal like, but you can always drag the values around. $\endgroup$ – J. W. Perry May 4 '14 at 6:30
  • $\begingroup$ I tried to add but the resulting line don't pass through the intersection. Can you please look here desmos.com/calculator/jvayrtjelf $\endgroup$ – Rustam Issabekov May 4 '14 at 6:38
  • $\begingroup$ It looks like lines y_1 and y_2 do have an intersection but the sum y_1 + y_2 does not pass through it. Can you please explain why? $\endgroup$ – Rustam Issabekov May 4 '14 at 6:39
  • $\begingroup$ Yeah it will not pass through the intersect of the two equations. What you have there is a third equation that has the sum of slopes of the first two equations, and the $y$ intercept is the sum of the $y$ intercepts of the two equations. That is what happens when you add the equations, versus when you set them equal to each other. $\endgroup$ – J. W. Perry May 4 '14 at 6:40
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Adding two linear equations which share some solution set (i.e., there are values for the variables which solve both) will result in a linear equation that also shares that solution set. This doesn't necessarily result in a "line" because linear equations only represent lines in 2 dimensions. In 3 dimensions they represent planes, and in higher dimensions they represent higher dimensional analogues.

A proper linear algebra course will teach you that adding and subtracting equations, or preforming row operations on the corresponding matrix, does not change the solution set to the system, for which the 2 dimensional analogue is lines intersecting as you mentioned.

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  • $\begingroup$ Can you please look at this graph desmos.com/calculator/jvayrtjelf $\endgroup$ – Rustam Issabekov May 4 '14 at 6:54
  • $\begingroup$ y_1 and y_2 do have an intersection but sum y_1 + y_2 does not intersect it. $\endgroup$ – Rustam Issabekov May 4 '14 at 6:55
  • $\begingroup$ You have not added the equations, you have added the values of $y_1$ and $y_2$. Do this by hand and you will see that $y_1 + y_2 = 2y = 8$. But to plot this correctly you want to have a single $y$ on one side, i.e., you want to plot $y = \frac{8}{2} = 4$, which still intersects the same point. $\endgroup$ – FlagCapper May 4 '14 at 23:31

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