MHB Why Does the Same y Value Satisfy Both Equations in Simultaneous Equations?

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In the discussion about simultaneous equations, the focus is on why the same y value satisfies both equations after determining x. The equations 2x + y = 10 and 3x + 4y = 25 yield x = 3 and y = 4, illustrating that a consistent linear system will always provide a single solution for both variables. The conversation highlights that if the equations are inconsistent, a contradiction arises immediately, indicating no solution exists. It also clarifies that linear systems can have zero, one, or infinitely many solutions, depending on their relationships graphically. Ultimately, finding a consistent solution confirms that both equations are satisfied simultaneously.
Poirot1
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consider the equations which we want to solve simultaneoulsy:

2x+y =10 (*)

3x+4y=25 (**)

By peforming 4(*) - (**), we have 5x =15 i.e. x=3. Now my question is this:

Why does the same y value (y=4) satisfy both original equations? This always happens. We never find that, having found x, the equations give 2 different values of y and we conclude there is no solution. In other words, if the equations are inconsistent, then the contradiction is derived immediately.
 
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Poirot said:
consider the equations which we want to solve simultaneoulsy:

2x+y =10 (*)

3x+4y=25 (**)

By peforming 4(*) - (**), we have 5x =15 i.e. x=3. Now my question is this:

Why does the same y value (y=4) satisfy both original equations? This always happens. We never find that, having found x, the equations give 2 different values of y and we conclude there is no solution. In other words, if the equations are inconsistent, then the contradiction is derived immediately.

This is a linear system of equations. Linear systems either have zero solutions, one solution, or infinitely many solutions. You can have linear systems where multiple values of $y$ solve the system: that's not no solution, it's infinitely many solutions.

Zero solution system:
$$\begin{align*}
3x+2y&=4\\
3x+2y&=2
\end{align*}$$

One solution system: your system.

Infinitely many solutions system:
$$\begin{align*}
3x+2y&=4\\
6x+4y&=8
\end{align*}.$$

In the first system, two identical LHS's equal two different numbers, which can never happen. Graphically, you can think of two parallel straight lines - no intersection. That's the contradiction you mentioned. In the second system, there's one and only one solution. Graphically, think of it as two non-parallel lines intersecting at one point. Finally, you have the third system, where the two equations are really saying the same thing: there's no new information in the second equation. Graphically, they are the same line, so they intersect everywhere.
 
Hello, Poirot!

I think you're over-thinking the problem . . .


Consider the equations which we want to solve simultaneoulsy:
. . 2x+y = 10 (*)
. . 3x+4y = 25 (**)

By peforming 4(*) - (**), we have: 5x =15 .---> .x = 3.

Now my question is this:
Why does the same y value (y = 4) satisfy both original equations?
This always happens. .It's supposed to!
Think of what it means to "solve the system".

We are to find values of $x$ and $y$ which satisfy both equations.

If $x = 3,\,y=4$ are correct, then of course they satisfy both equations.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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