Answer:
We have two relations
2*x*y = 6 (2 times x times y is equal to six)
x - y = 3 ( the difference between x and y is trhe)
Where whe have two variables, x and y.
To solve this system, the first step is isolation one of the variables in one of the equations, let's isolate x in the second equation.
x - y = 3
x = 3 + y
now we can replace this in the other equation and then solve it for y.
2*x*y = 6
2*(3 + y)*y = 6
6y + 2y^2 = 6
now we have the quadratic equation:
2y^2 + 6y - 6 = 0
the solutions are:
[tex]y = \frac{-6 + -\sqrt{6^2 - 4*2*(-6)} }{2*2} = \frac{-6 +- 9.2}{4}[/tex]
the solutions are:
y = (-6 + 9,2)/4 = 0.8
y = (-6 - 9.2)/4 = -3.8
if y = 0.8, then:
x = 3 + y = 3.8
if y = -3.8
x = 3 + y = 3 - 3.8 = -0.8
so we have two possible solutions:
(-0.8, -3.8) and (3.8, 0.8)
