Try plugging those numbers you get into your equations to see whether they're satisfied or not.
Show us what you're doing.
I have the following problem to solve:
Solve these equations for both v and x:
3v + 4 x = 3,
7v - 3 x = 3
The way I learned to set it up is to isolate a variable and then plug the solution into the other equation, but when I try to do so I do not get nice numbers that don't seem to be correct.
It is much easier to solve equations by using the method of elimination by equating coefficient.
We may eliminate any one variable by equating its coefficient and then adding / subtracting the two equations.
Let us consider the given equations:
3v + 4 x = 3 ------ (1)
7v - 3 x = 3 --------(2)
let us eliminate x, for that we will equate coefficient of x in both the equations by multiplying the equqtion with suitable number, in this case we will multiply the equation (1) by 3 and equation (2) by 4, we get:
9v + 12 x = 9 ------ ( 3 )
28v - 12 x = 12------( 4)
Since sign of the term containing x is different we add equation ( 3 ) and ( 4 ), we get
37 v = 21 That gives v = 21/37 Now you can get the value of x by plugging in this value in any equation ( 1 ) or ( 2)
The error I made was in my algebra. When solving the first equation for v, I got stuck at this point:
v = (3 - 4x) / (3)
I did not take into consideration that the denominator of 3 affects both terms in the numerator and simply canceled the 3 within numerator & denominator, failing to apply it to the 4x term as well. That is why my solution didn't make sense when continuing the problem from thereon.
Thank you for the visual display, it certainly helped! And I now know what I did wrong.
I am going to apply your method to the rest of my problems and see which method fits better for me. The first method is more intuitive for me, though.