# Simultaneous Equations

#### Poirot

##### Banned
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
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*}

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
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.