How to solve 3 simultaneous equation?

[Firoj]

It's been a while since I've had to do a simultaneous equation and I'm rusty on a few of the particulars.

Say for example I have the following equations:

x + y = 7
2x + y + 3z = 32
2y + z = 13
I know that I need to combine the above 3 equations into 2 other equations, for example, if I combine (counting down) 1 + 2 I'd get

3x + 2y + 3z = 32
And combing 2 with 3 I'd get

2x + 3y + 4z = 45
Which is fine, and I understand. It's the next steps I have trouble understanding. A lot of the examples I've been looking at have a value for each of the x, y z. Looking at this site here I'm not sure what is going on step 2.

I can see that they are multiplying one line by 2. Is that something you always do? Like, with simultaneous equations do you always multiple one of the equations by 2? If not, how do you determine which number to use?

My understanding of simultaneous equations is extremely limited.

[bilu96]

This is a difficult problem to say the least

[HallsofIvy]

You are misunderstanding!

It is NOT a matter of just "combine the above 3 equations into 2 other equations", you combine the equations is such a way that you also eliminate one of the unknowns, reducing two equations in two unknowns, then combine those to get a single equation in one unkown.
'
"for example, if I combine (counting down) 1 + 2 I'd get
3x + 2y + 3z = 32
And combing 2 with 3 I'd get
2x + 3y + 4z = 45
Which is fine, and I understand."

No, you don't understand. Yes, you can do those things but why? It doesn't help!

Instead, starting with the three equations
x + y = 7
2x + y + 3z = 32
2y + z = 13'

We see that the first equation, x+ y= 7, does not involve z we might decide to use the other two equations to eliminate z. One way is to rewrite 2y+ z= 13 as z= 13- 2y so that 2x+ y+ 3z= 2x+ y+ 3(13- 2y)= 2x+ y+ 39- 6y= 2x- 5y+ 39= 32.

Subtract 39 from both sides to get 2x- 5y= -7. Now use that and x+ y= 7 to eliminate either x or y. We can write y= 7- x so 2x- 5y= 2x- 5(7- x)= 2x- 35- 5x= -7 so -3x= 27. Solve that for x then use that value to reduce the other equations.

[Guest]

[tex]\begin{array}{|l} x + y = 7 \\ 2x + y+3z = 32\\2y+z=13 \end{array}[/tex]

[tex]\begin{array}{|l} x = 7-y \\ 2x+y+3z = 32\\z =13-2y\end{array}[/tex]

2(7-y)+y+3(13-2y)=32

14-2y+y+39-6y=32

-7y=-21 [tex]\Rightarrow[/tex] y=3

[tex]\begin{array}{|l} x = 7-y \\ z = 13-2y \end{array}[/tex]

[tex]\begin{array}{|l} x = 7-3 \\ z = 13-2.3 \end{array}[/tex]

(4;3;7)

[Firoj]

There is so much content online and it's hard to know where to start. Normally, hallway conversations help me.

[HallsofIvy]

Well, that depends on which hallway! I've known some hallways I wouldn't even want to walk down!