x.y=30 and x+y=10.3 ,what is x & Y.

final answer is x=5.15+1.86i
and y=5.15-186i, this way is by try and error (logic). but can somebody please show me how you do it by aljebra, please, thank you.

xy=30

x+y=10.3

30/y + y=10.3

30+y^2=10.3y

y^2-10.3y+30=0
quadratic equation.

y= (10.3 +-sqrt(10.3^2 -120))/2

solve that.

that is the problem i cant solve it because i will end up with square root of -13.91. please solve this equation for me.

Goodness. I will do the sqrt for you.

sqrt(10.3^2-120)=sqrt(-13.91)=
=sqrt(13.91)(sqrt-1)=sqrt(13.91)i
=3.72i

SO you have:
(10.3+-3.72i)/2=5.15+-1.86i

very good, but when i want to proove it by x + y =10.3,

x=30/(5.15+1.84i), y=(5.15+1.84i)
how you drive this equation
30/(5.15+1.84i)+(5.15+1.84i)=10.3
thank you

To solve the given system of equations algebraically, we can use substitution or elimination method. Let's use the substitution method:

Given:
Equation 1: x * y = 30
Equation 2: x + y = 10.3

We can rearrange Equation 2 to express x in terms of y. Subtracting y from both sides, we have:
x = 10.3 - y

Substitute this expression for x in Equation 1:
(10.3 - y) * y = 30

Expanding this equation, we have:
10.3y - y^2 = 30

Rearrange the equation to a quadratic form:
y^2 - 10.3y + 30 = 0

Now, we can solve this quadratic equation for y. We can either factor it or use the quadratic formula.

Using the factoring method:
(y - 5.15)(y - 5.85) = 0

From this equation, we have two possibilities for y:
1) y - 5.15 = 0 --> y = 5.15
2) y - 5.85 = 0 --> y = 5.85

Now that we have found the possible values of y, we can substitute them back into Equation 2 to find the corresponding values of x.

For y = 5.15:
x + 5.15 = 10.3
x = 10.3 - 5.15
x = 5.15

For y = 5.85:
x + 5.85 = 10.3
x = 10.3 - 5.85
x = 4.45

Therefore, the solutions to the system of equations are:
x = 5.15 and y = 5.15 (or)
x = 4.45 and y = 5.85