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Simultaneous Equations

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Poirot

Banned
Feb 15, 2012
250
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.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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.
 

soroban

Well-known member
Feb 2, 2012
409
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.