SOLUTION: xy=30 3x+y=1

Question 1153062: xy=30
3x+y=1
Found 4 solutions by MathLover1, Alan3354, josgarithmetic, greenestamps:
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!
...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)   (Show Source): You can put this solution on YOUR website!
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)   (Show Source): You can put this solution on YOUR website!
-------------
xy=30
3x+y=1
-------------

-

Using real number coordinates, no intersection in the plane.

Answer by greenestamps(13200)   (Show Source): You can 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.

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