Answer :
[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)
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)
