SOLUTION: You must show all of your work to receive credit. All or nothing. Solve the system using any method. 3/x+4/y=5/2. 6/x+1/y=1/2

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: You must show all of your work to receive credit. All or nothing. Solve the system using any method. 3/x+4/y=5/2. 6/x+1/y=1/2      Log On


   



Question 1093252: You must show all of your work to receive credit. All or nothing. Solve the system using any method. 3/x+4/y=5/2. 6/x+1/y=1/2
Found 3 solutions by josgarithmetic, MathTherapy, ikleyn:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
system%283%2Fx%2B4%2Fy=5%2F2%2C6%2Fx%2B1%2Fy=1%2F2%29

The second equation gives you 1%2Fy=1%2F2-6%2Fx;
Substitute into first equation to form 3%2Fx%2B4%28%28x-12%29%2F%282x%29%29=5%2F2.
Simplify and solve this....
x=-42
.
.
.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
You must show all of your work to receive credit. All or nothing. Solve the system using any method. 3/x+4/y=5/2. 6/x+1/y=1/2
3%2Fx+%2B+4%2Fy+=+5%2F2%09 ------ eq (i)
6%2Fx+%2B+1%2Fy+=+1%2F2 ------ eq (ii)	
-+6%2Fx+-+8%2Fy+=+-+10%2F2 ------ Multiplying eq (i) by - 2 ------ eq (iii)	
-+7%2Fy+=+-+9%2F2 ---------- Adding eqs (iii) & (ii)	
- 9y = - 14 -------- Cross-multiplying


============================

Instead of substituting the value of y into one of the equations, it's best to repeat the process, as follows:
3%2Fx+%2B+4%2Fy+=+5%2F2%09 ------ eq (i)
6%2Fx+%2B+1%2Fy+=+1%2F2 ------ eq (ii)	
-+24%2Fx+-+4%2Fy+=+-+4%2F2 ------ Multiplying eq (ii) by - 4 ------ eq (iii)	
-+21%2Fx+=+1%2F2 ---------- Adding eqs (iii) & (i)	
x = - 21(2) ------- Cross-multiplying
highlight_green%28matrix%281%2C3%2C+x%2C+%22=%22%2C+-+42%29%29

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
There is a standard method to solve such non-linear systems.

Introduce new variables  u = 1/x  and  v = 1/y.

Then the original system of non-linear equations becomes (by a magical way) a linear system relative new variables "u" and "v".

3u + 4v = 5/2,   (1)
6u +  v = 1/2.   (2)


Now you can apply any method you want / you know to solve (1), (2).

I will use Elimination method. 

Multiply equation (1) by 4 (both sides), and equation (2) by 2 (both sides).


You will get

12u + 16v = 10,   (3)
12u +  2v = 1.    (4)


Subtract eq(4) from eq(3) to get

14v = 9.

Then  v = 9/14.

Having this, you easily get (from (4))  12u = 1 - 2*(9/14) = 1 - 18/14 = -4/14 = -2/7.

Hence,  u = -2/(7*12) = -1/42.

Thus you get  u = -1/42,  v = 9/14.


Now recall how you introduced u and v:  u = 1/x = -1/42;  hence,  x = -42.

                                   and   v = 1/y = 9/14;  hence,  y = 14/9.

The original system solved.  The solutions are  x = -42  and  y = 14/9.


You can substitute it into the original equations to check it on your own (I did it).


-------------
To see more tricks of this kind,  look into the lesson
    - Solving systems of non-linear equations in two unknowns using the Cramer's rule
in this site.


Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic
     "2x2-Matrices, determinants, Cramer's rule for systems in two unknowns"


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.