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I know, it is very simple, but I'm get stuck. Let I have two equations of straight lines on a plate.

For example: $2x + 3y + 5 = 0$; 3$x + 4y + 7 = 0$

What is the geometric meaning of a new line, that has been got by equaling of equations.

$2x + 3y + 5 = 3x + 4y + 7$;

$x + y + 2 = 0$

It is a line, lies exactly between them, isn't it?

I have a spirit of an old lecture repeats in my mind. "To find an intersection point of two functions you need to equal the equations..."

But if I let x = any constant and $y = - 2 - x$ , then I'll get a complete rubbish.

What do I forget?

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up vote 6 down vote accepted

It is a line through the point of intersection of the two given lines. One can say no more than that.

Note for example that $2x+3y+5=0$ and $9x+12y+21=0$ gives the same set of two lines as your two equations. But if we set $2x+3y+5=9x+12y+21$, we get a very different line from $x+y+2=0$. The only shared feature is that it goes through the point of intersection of our two original lines.

Added: The question has been edited, and in addition it is asked how to find the point of intersection of $2x+3y+5=0$ and $3x+4y+7=0$. The idea is to eliminate one of the variables.

The first equation is equivalent to $8x+12y+20=0$. (We multiplied through by $4$.)

The second equation is equivalent to $9x+12y+21=0$.

If the two equations hold, then $8x+12y+20=9x+12y+21$. The $12y$ "cancel," and we get $x=-1$. Then from $2x+3y=5$ we get $y=-1$.

Usually this is put in a somewhat different way. For example, multiply the first equation through by $4$, and the second by $-3$. We get $8x+12y+20=0$ and $-9x-12y-21=0$. Add. We get $-x+-1=0$, so $x=-1$.

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No, it is not the line that lies in between the two lines.

x+y+2=0 is the line that that passes through the point of intersection of your two given lines.

You can see this because setting the two equations equal is equivalent to asking where the point of intersection.

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