SOLUTION: please help me...how to solve these simultaneous eguations... (4/a + b) + ( 2/a-b) = 2 ; (2/a+b) + (4/a-b) = 5/2 and one more query.. can you please tell me if this equation ha

Click here to see ALL problems on Linear Equations And Systems Word Problems

Question 762701: please help me...how to solve these simultaneous eguations...
(4/a + b) + ( 2/a-b) = 2 ; (2/a+b) + (4/a-b) = 5/2
and one more query..
can you please tell me if this equation has unique solution,infinitely many solutions or no solution...
x - 2y = 3 ; 3x/2 -y = 1
I would be highly obliged for your help.. thanking you
waiting for your earliest reply
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
please help me...how to solve these simultaneous eguations...
(4/a + b) + ( 2/a-b) = 2
(2/a+b) + (4/a-b) = 5/2
------
Multiply thru the top equation by 2 to get:
(8/a+b) + (4/a-b) = 4
(2/a+b) + (4/a-b) = 5/2
------------------
Subtract to get:
(6/a+b) = 3/2
a+b = 4
=====================
Substitute and solve for "a+b" and solve for "a-b":
(2/4) + (4/a-b) = (5/2)
----------
4/a-b = 2
a-b = 2
---------------------
Now, solve for a and for b:
a + b = 4
a - b = 2
----
Add to solve for "a:
2a = 6
a = 3
----
Solve for "b":
a + b = 4
3 + b = 4
b = 1
====================
can you please tell me if this equation has unique solution,infinitely many solutions or no solution...
x - 2y = 3
3x/2 -y = 1
-------------------
Solve for "y":
y = (3/2)x - 1
-------
Substitute for "y" and solve for "x":
x - 2[(3/2)x] = 3
x -3x = 3
x = -3/2
-----
Solve for "y":
y = (3/2)x - 1
y = (3/2)(-3/2) - 1
y = -2
=========================
Ans: One solution
=========================
Cheers,
Stan H.
=========

Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

This one was very interesting. Going at it straight up gets you into trouble. But I noted that the denominators in each of the two equations were the same, so I tried:

Let and let . Making the substitutions, the two equations become:

Solve by elimination to get

Then

Solve the simple 2X2 by elimination.

John

Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it