SOLUTION: xy=30 3x+y=1
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Question 1153062
:
xy=30
3x+y=1
Found 4 solutions by
MathLover1, Alan3354, josgarithmetic, greenestamps
:
Answer by
MathLover1(20849)
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...eq.1
...eq.2
-----------------------------
...eq.1, solve for
.........eq.1a
substitute in eq.2
...eq.2
solutions:
or
go to
.........eq.1a, substitute
or
Answer by
Alan3354(69443)
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xy=30
3x+y=1
=================
y = 30/x
--
3x + 30/x = 1
3x^2 + 30 = x
3x^2 - x + 30 = 0
Solved by
pluggable
solver:
SOLVE quadratic equation (work shown, graph etc)
Quadratic equation
(in our case
) has the following solutons:
For these solutions to exist, the
discriminant
should not be a negative number.
First, we need to compute the discriminant
:
.
The discriminant -359 is less than zero. That means that there are no solutions among real numbers.
If you are a student of advanced school algebra and are aware about
imaginary numbers
, read on.
In the field of imaginary numbers, the square root of -359 is + or -
.
The solution is
, or
Here's your graph:
Answer by
josgarithmetic(39617)
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-------------
xy=30
3x+y=1
-------------
-
Using real number coordinates, no intersection in the plane.
Answer by
greenestamps(13200)
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put this solution on YOUR website!
If you are looking for solutions in real numbers, it is easy to quickly determine that there are none through logical reasoning.
(1) The equation
tells us that x and y are either both positive or both negative.
(2) Knowing that, the equation
tells us that x and y are both positive, and that both are less than 1.
(3) Logical reasoning then tells us that it is not possible to have a product of 30 using two numbers less than 1.
ANSWER: No solution in real numbers.