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Elimination:
3x-4y=16
5x+6y=14


Sagot :

[tex] \left \{ {{3x-4y=16\ \ | * (-5)} \atop {5x+6y=14 \ \ | *3}} \right. \\\\ \left \{ {{-15x+20y=-80} \atop {15x+18y=42}} \right. \\+-----\\add\ both \ equations\\\\ 38y=-38\ \ \ | divide\ by\ 38\\\\ y=-1\\\\ 3x=16+4y\\\\ x=\frac{16+4y}{3}=\frac{16+4*(-1)}{3}=\frac{12}{3}=4\\\\ \left \{ {{y=-1} \atop {x=4}} \right. [/tex]
okay for this one, i will start with x

3x - 4y = 16
5x + 6y = 14

I will multiply the top row with (-5) and the bottom with (3) 
so the new equation is

-15x + 20y = -80
15x + 18y = 42

After canceling (x) the new equation will be

38y = -38

Divide by 38

And you will get -1

Now substitute (-1) to (y)  to the old equation ( any of the old equation)
3x- 4(-1)= 16
x = 4

answers ( 4, -1)