# Math help

How would you solve these systems by elimination? I'm in desperate need of help!

1. -35x-15y=0
-56x+24y=0

2. -56x+72y=-16
-49x+63y=-14

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1. we must find a common multiple of something here
15 = 3*5
24 = 3*8
so I guess we could use 3*5*8 = 120

15 * 8 = 120 so multiply first eqn by 8
24 * 5 = 120 so multiply second by 5
- 280 x - 120 y = 0
- 200 x + 120 y = 0
-480 x = 0
so x = 0
and y = 0
which we could have said back at the beginning if we had sketched graphs

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posted by Damon
2. in the second one both 56 and 49 have common factor 7 which we will take advantage of
56 = 7 * 8
49 = 7*7 so use 7*7*8 = 392

56 *7 = 392 so multiply first by 7
49 *8 = 392 so multiply second times 8

I think you can take it from there.

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posted by Damon
3. On the first one how did you get -280 and -200?

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4. -280 = -35 * 8

-280 =-56 * 5
sorry, 280 not 200, does not matter, solution is origin (0,0) anyway
both of those lines are through the origin having no b in y = m x + b

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posted by Damon

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