Hello , just wanted to talk about solving equations fastly and precisely

i can't devolope a techinque !!!

Let's Have 2 Equations

5 = (v1-v2)/15 + v1/5

2 = (v2-v1)/15 +v2/4



7 = v1/5 + v2/4 => 140 = 4v1 + 5v2

5v2 = 140 - 4v1 => v2 = (140-4v1)/5

140 = 4v1 + 140-4v1 , 0=0 !!!!

!!! , how do i avoid this ?

please guide me to a good way of thinking to avoid all the traps .

i know that i'm noob , please help me .

 

Hello , just wanted to talk about solving equations fastly and precisely

i can't devolope a techinque !!!

Let's Have 2 Equations

5 = (v1-v2)/15 + v1/5

2 = (v2-v1)/15 +v2/4



7 = v1/5 + v2/4 => 140 = 4v1 + 5v2

5v2 = 140 - 4v1 => v2 = (140-4v1)/5

140 = 4v1 + 140-4v1 , 0=0 !!!!

!!! , how do i avoid this ?

please guide me to a good way of thinking to avoid all the traps .

i know that i'm noob , please help me .

The problem with your approach is that you tried to solve the problem with only one equation, even though you have two unknowns. When you add two equations together, you do not retain all relevant information from both equations. The new equation is valid, but only qualifies as one equation. With two unknowns, you need to use two independent equations.

Your first step is actually a good one because you took your two equations and combined them into one equation that ended up being simpler than either of the first two equations. However, you still need to use one more equation. You could use this new equation with either of the first two equations, or you could subtract the first two equations (rather than add them, as you did the first time) to get a new equation. You will then have two independent equations to solve.

As far as solving the two equations that you finally choose, there are many techniques. The most straightforward would be to take one equation and represent V1 in terms of V2 (that is V1=f(V2)), then substitute that relation into the other equation. That new equation will only have V2 as an unknown, and you can solve for V2. Then, V1 is easily solved with your "substitution" equation V1=f(V2).

 

Another way that works for me is to solve the first equation for a variable, say V1. Then substitute that solution for V1 in the second equation. Once you have a number for v2, V1 is easy...

 

I successfully solved the two linear equations and obtained workable values for V1 and V2.

That leads me to believe that the two equations are independent.

I suggest you review linear algebra techniques for solving two equations with two unknowns.

hgmjr

 

i am helped , thanks !! .