SOLUTION: Ok so we are doing these problems where you have to find x and y and the equations are 4x+3y=9 and the other one is 3x+4y=12 and what I did was I multiplied 4x+3y=9 and -3 and then

Algebra ->  Expressions-with-variables -> SOLUTION: Ok so we are doing these problems where you have to find x and y and the equations are 4x+3y=9 and the other one is 3x+4y=12 and what I did was I multiplied 4x+3y=9 and -3 and then      Log On


   

Question 577732: Ok so we are doing these problems where you have to find x and y and the equations are 4x+3y=9 and the other one is 3x+4y=12 and what I did was I multiplied 4x+3y=9 and -3 and then i multiplied 3x+4y=12 and 4. But my question is would you multiply the 9 and 12 to? Or would you just multiply the equations? And did I multply by the right numbers?
Answer by KMST(5398) About Me  (Show Source):
You can put this solution on YOUR website!
So you have the system of linear equations
system%284x%2B3y=9%2C3x%2B4y=12%29
and you are trying to solve it by "elimination."
4x%2B3y=9 is equivalent to %28-3%29%284x%2B3y%29=%28-3%29%2A9 --> -12x-9y=-27
If the expression 4x%2B3y is equal to 9,
the value of (-3) times that expression will be (-3) times 9.
Multiplying both sides of the equation by the same non-zero number, gives you another equivalent equation, with exactly the same solutions.
3x%2B4y=12 is equivalent to 4%283x%2B4y%29=4%2A12 --> 12x%2B16y=48
If you did those transformations and got the equivalent equations
-12x-9y=-27 and 12x%2B16y=48%29,
you are on the right path.
Next you would add those equations and keep the resulting equation
-12x-9y%2B12x%2B16y=-27%2B48%29 --> 7y=21 --> highlight%28y=3%29
along with one of your original equations:
system%28y=3%2C4x%2B3y=9%29
At that point, you would substitute the value of y into the second equation and find that highlight%28x=0%29