SOLUTION: 2/x+5/y=5/6; x/2+y/5=5

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Question 1132110: 2/x+5/y=5/6; x/2+y/5=5
Found 3 solutions by ikleyn, MathLover1, MathTherapy:
Answer by ikleyn(52777) About Me  (Show Source):
You can put this solution on YOUR website!
.
2%2Fx + 5%2Fy = 5%2F6,         (1) 

x/2 + y/5 = 5.       (2)


This system of equations in non-linear.

To make the solution easier, I will introduce NEW VARIABLES  a = 2%2Fx,  b = 5%2Fy.  Then the system takes the form


a + b = 5%2F6          (3)

1%2Fa + 1%2Fb = 5         (4)


To solve it, first simplify equation (4):


    1%2Fa + 1%2Fb = 5  ====> %28a%2Bb%29%2F%28ab%29 = 5  ====>  replace a+b by 5%2F6, based on (3)  ====>  %28%285%2F6%29%29%2F%28ab%29 = 5  ====>  ab = 1%2F6.


Next, from (3), express b = 5%2F6 - a  and substitute it into equation  ab = 1%2F6.  You will get


    a%2A%285%2F6-a%29 = 1%2F6,

    -6a^2 + 5a = 1

    6a^2 - 5a + 1 = 0.


Solve the last quadratic equation using the quadratic formula.

You will get two solutions:  a= 1%2F2  and  a= 1%2F3.


Thus, the system  (3)-(4) has two solutions:


    1)  a= 1%2F2,  b= 5%2F6 - 1%2F2 = 2%2F6 = 1%2F3,     

and

    2)  a = 1%2F3,  b= 5%2F6 - 1%2F3 = 5%2F6+-+2%2F6 = 3%2F6 = 1%2F2.


Now you need to return from "a" and "b" to x and y, via the formulas  a = 2%2Fx,  b = 5%2Fy.


By doing so, you get two solutions for the original system:


1)  x = 2%2Fa = 2%2F%28%281%2F2%29%29 = 4;    y = 5%2Fb = 5%2F%28%281%2F3%29%29 = 15.


2)  x = 2%2Fa = 2%2F%28%281%2F3%29%29 = 6;    y = 5%2Fb = 5%2F%28%281%2F2%29%29 = 10.


Answer.  The system has two solutions  1)  x= 4; y= 15    and   2)  x= 6,  y= 10.

Solved.

You may check that the solution is correct by substituting the found values into the original equations.


----------------

I am very glad that the tutor @MathLover1 placed her solution here.

Comparing these two,  you can see how many tons of calculations I saved you from,  using my substitutions  !  !  !


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To see other similar solved problems for systems of two non-linear equations in two unknowns,  look into the lessons
    - Solving systems of non-linear equations by reducing to linear ones
    - Solving systems of non-linear equations in two unknowns using the Cramer's rule
in this site.


Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

2%2Fx%2B5%2Fy=5%2F6+........1)
x%2F2%2By%2F5=5+........2)
----------------------------------------------
2%2Fx%2B5%2Fy=5%2F6+........1).....both sides multiply by 6xy
%286xy%2A2%29%2Fx%2B%286xy%2A5%29%2Fy=%286xy%2A5%29%2F6+
12y%2B30x=5xy+........solve for y
12y-5xy=-30x+
%2812-5x%29y=-30x+
y=-30x%2F%2812-5x%29+........1a)

x%2F2%2By%2F5=5+........2)......both sides multiply by 10
10x%2F2%2B10y%2F5=10%2A5+
5x%2B2y=50+........solve for y
y=50%2F2-5x%2F2+
y=25-%285%2F2+%29x.....2a)

from 1a) and 2a) we have
-30x%2F%2812-5x%29+=25-%285%2F2+%29x.......solve for x
5x%2F2+-30x%2F%2812-5x%29+=25
%285x%2812-5x%29%29%2F2%2812-5x%29+-%282%2A30x%29%2F2%2812-5x%29+=25
%2860x-25x%5E2+-60x%29%2F2%2812-5x%29+=25
-25x%5E2+=25%2A2%2812-5x%29..........simplify
-x%5E2+=2%2812-5x%29
-x%5E2+=24-10x
0+=x%5E2-10x%2B24....factor
%28x+-+4%29+%28x+-+6%29=0
=> x=4 or x=6
go to y=25-%285%2F2+%29x.....2a), substitute x
y=25-%285%2F2+%294
y=25-5%2A2+
y=25-10+
y=15+
y=25-%285%2F2+%296
y=25-5%2A3+
y=25-15+
y=10+

solutions:
x=4,y=15+

or
x=6,y=10+






Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!
2/x+5/y=5/6; x/2+y/5=5
matrix%281%2C3%2C+2%2Fx+%2B+5%2Fy%2C+%22=%22%2C+5%2F6%29
12y + 30x = 5xy ------ Multiplying by LCD, 6xy ------ eq (i)

matrix%281%2C3%2C+x%2F2+%2B+y%2F5%2C+%22=%22%2C+5%29
5x + 2y = 50 --------- Multiplying by LCD, 10
5x = 50 - 2y --------- eq (ii)

12y + 30x = 5xy =====> 12y + 6(5x) = y(5x)
12y + 6(50 - 2y) = y(50 - 2y) ------- Substituting 50 - 2y for 5x in eq (i)
matrix%281%2C3%2C+12y+%2B+300+-+12y%2C+%22=%22%2C+50y+-+2y%5E2%29
matrix%281%2C3%2C+2y%5E2+-+50y+%2B+300%2C+%22=%22%2C+0%29
matrix%281%2C3%2C+2%28y%5E2+-+25y+%2B+150%29%2C+%22=%22%2C+2%280%29%29
matrix%281%2C3%2C+y%5E2+-+25y+%2B+150%2C+%22=%22%2C+0%29
(y - 15)(y - 10) = 0
y = 15             OR           y = 10

5x = 50 - 2(15) ------- Substituting 15 for y in eq (ii)
5x = 20
matrix%281%2C5%2C+x%2C+%22=%22%2C+20%2F5%2C+%22=%22%2C+4%29


5x = 50 - 2(10) ------- Substituting 10 for y in eq (ii)
5x = 30
matrix%281%2C5%2C+x%2C+%22=%22%2C+30%2F5%2C+%22=%22%2C+6%29