SOLUTION: "solve each system by elimination, first clear denominators" (x+6)/5 + (2y-x)/10 = 1 (x+2)/4 + (3y+2)/5 = -3 thanks!

Algebra ->  Algebra  -> Linear-equations -> SOLUTION: "solve each system by elimination, first clear denominators" (x+6)/5 + (2y-x)/10 = 1 (x+2)/4 + (3y+2)/5 = -3 thanks!      Log On

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Question 24663: "solve each system by elimination, first clear denominators"
(x+6)/5 + (2y-x)/10 = 1
(x+2)/4 + (3y+2)/5 = -3
thanks!
Answer by algebrapro18(206) About Me  (Show Source):
You can put this solution on YOUR website!
(x+6)/5 + (2y-x)/10 = 1
(x+2)/4 + (3y+2)/5 = -3

to clear the fractions from the first equation all you have to do is multiply the entire equation by 10. you will end up with:

2(x+6)+(2y-x)=10

simplifying you get

2x+12+2y-x=10
x+2y+12=10
x+2y=-2

to clear the fractions from the secound equation all you have to do is multiply the entire equation by 20. you will end up with

5(x+2)+4(3y+2)=-60

simplifying you get

5x+10+12y+8=-60
5x+12y+18=-60
5x+12y=-78

now we multiply equation 1 by -5 and add the 2 equations together

-5(x+2y=-2)
-5x-10y=10

5x+12y=-78
-5x-10y=10
============
2y=-68
y=-34


now pluging that into equation 2

5x+12(-34)=-78
5x-408=-78
5x=330
x=66

and that is the answer (66,-34)