So, we can write it like this:
x+y=3
x*y=12
from the first we know that
x+y=3
y=3-x
so:
x*y=12 can be written as:
x*(x-3)=12
so x^2-3x=12
Actually this equation does not have a solution within the real numbers
but we can calculate:
x^2-3x-12=0
and if we use the quadratic formula, we will have:
(remember the quadratic formula is the following, where for us a=1,b=-3 and c=-12)
So this would be the solution, where the other number would be
The other solution would be the following one;
x=
, y=
x+y=3
x*y=12
from the first we know that
x+y=3
y=3-x
so:
x*y=12 can be written as:
x*(x-3)=12
so x^2-3x=12
Actually this equation does not have a solution within the real numbers
but we can calculate:
x^2-3x-12=0
and if we use the quadratic formula, we will have:
(remember the quadratic formula is the following, where for us a=1,b=-3 and c=-12)
So this would be the solution, where the other number would be
The other solution would be the following one;
x=