So we have this system of two equations in two unknowns:
Solve the second one for y by subtracting x from both sides:
Substitute (5-x) for y in x∙y = -10
Get 0 on the right by adding 10 to both sides of the equation:
Get the left side in descending order:
Multiply through by -1 to make the leading term positive:
(Change all the signs):
That doesn't factor so we must use the quadratic formula:
a=1, b=-5, c=-10
So there are two solutions for x:
and
We find y for each solution using y=5-x
and and and and and
There really aren't two solutions, it's just a matter of which one
we call x and which one we call y.
Answer: and
Checking:
Multiply them:
<-- that checks!
Add them:
<-- that also checks!
Edwin
You can put this solution on YOUR website! I assume that x and y are Real numbers
:
1) x + y = 5
:
2) xy = -10
:
solve equation 1 for x and substitute in equation 2
:
x = 5 - y
:
(5-y)y = -10
:
5y -y^2 = -10
:
y^2 -5y -10 = 0
:
use quadratic formula
:
y = (-(-5) + square root((-5)^2 - 4 * 1 * (-10)))/(2*1)) = 6.5311
:
y = (-(-5) - square root((-5)^2 - 4 * 1 * (-10)))/(2*1)) = -1.5333
:
there are two solutions
:
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if y = 6.5311, x = 5 - 6.5311 = -1.5311
:
if y = -1.5311, x = 5 - (-1.5311) = 6.5311
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