what is the solution of the following system of equation? 2x-6=2y 3-2x=y

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what is the solution of the following system of equation? 2x-6=2y 3-2x=y

Mathematics
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x=2 and y=-1
plug in 3-2x for y for the equation 2x-6
right, substitute: you could do as bradley said: y = (3 - 2x) --> (2x - 6) = 2*(3 - 2x) --> (2x-6) = (6 - 4x)

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so 2x + 4x -6 = 6, therefore 6x - 6 = 6, therefore x = 12/6 = 2, x = 2, now use the fact that 3-2x=y --> y now is 3-2*(2) = 3-4 = -1 --> (x,y) = ( 2, -1 )
ok. i am going to see if i can understand it so i can work on my next one
so, just tell me how i sovled it?
so you get the general concept
i am confused
ok, what is confusing now?
i guess the way it is written
ok, you have two equations and two variables
toussainr might be able to explain it differently...
ok, well then: the goal is find the values for each variable that satisfy both equations
My browser just froze, sorry
dpflan, please go ahead!!!
okay, I'm going to give it try, hoping I'm helpful
you have to equations. we are going to proceed in steps to make things easier!!!
1. You move terms in 'x' and 'y' to the left hand side (LHS) and the constants to the right hand side (RHS)
So we have: 2x-2y = 6
AND
-2x-y = -3
Let's move on with step two
2. From equation "-2x-y = -3" Let's derive a value for 'y'
We have -y = -3+2x OR y = 3-2x (we multiply by -1 the LHS and the RHS)
This value: "y = 3-2x" we are going to replace it in the first equation : 2x-2y = 6
We have now:
2x - 2 (3-2x) = 6
Let's multiply,. we have: 2x - 6 + 4x = 6
Let's perform the required addition: 6x-6 = 6
let's divide by 6 the RHS and the LHS, we have now : "x - 1 = 1"
Form here we can easily find the value of x :" x = 2"
Recall that y = 3-2x So, we replace: y = 3 - 2 * 2 ( * stands here for multiplication ) And finally y = 3 -4 y = -1
THAT"S THE BEST I CAN DO!
thank you. I understand it more now. :)
Okay!!! Just to quote John Von Neuman : "You don't understand maths, you get accustomed to it" So do a many exercises as you can! That's the key to successful mathematics!

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