SOLUTION: (1)/(3)X + (1)/(5)Y =7 (1)/(6)X - (2)/(5)Y = -4 using elimination method.

Question 701038: (1)/(3)X + (1)/(5)Y =7
(1)/(6)X - (2)/(5)Y = -4
using elimination method.
Found 2 solutions by Alan3354, checkley79:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
(1)/(3)X + (1)/(5)Y =7
(1)/(6)X - (2)/(5)Y = -4
using elimination method.
---------------
Your parentheses don't contribute anything.
1/3X + 1/5Y =7
1/6X - 2/5Y = -4
------------
I'll assume you mean:
(1/3)X + (1/5)Y =7
(1/6)X - (2/5)Y = -4
See the difference?
It clarifies that the x & y is not in the denominator.
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(1/3)X + (1/5)Y = 7 times 15 --> 5x + 3y = 105
(1/6)X - (2/5)Y = -4 time 30 --> 5x -12y = -120
-------
5x + 3y = 105
5x -12y = -120
-------------- Subtract
15y = 225
y = 15
----
x = 12
Answer by checkley79(3341) (Show Source): You can put this solution on YOUR website!
(1)/(3)X+(1)/(5)Y=7 MULTIPLY BY 2 & ADD.
(1)/(6)X-(2)/(5)Y=-4
(2)/(3)X+(2)/(5)Y=14 ADD.
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(1/6+2/3)X=14-4
(1/6+4/6)X=10
(5/6)X=10
5X/6=10 CROSS MULTIPLY
5X=10*6
5X=60
X=60/5
X=12 ANS.
(1/3)12+(1/5)Y=7
4+Y/5=7
Y/5=7-4
Y/5=3
Y=15 ANS.
PROOF:
(1)/(3)*12+(1)/(5)*15=7
4+3=7
7=7

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