решити методом алгебраического сложения:
1) 2х-3у=1
3х+у=7
2) х+у=7
5х-7у=11
3)8х+3у=-21
4х+5у=-7
1) 2х-3у=1
3х+у=7
2) х+у=7
5х-7у=11
3)8х+3у=-21
4х+5у=-7
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Answers & Comments
3x+y=7 3x+y=7·3 9x+3y=21 2x-3y+9x+3y=22
5) 11x=22 6) 4-3y=1
x=2 y=1 (2;1)
1)x+y=7
5x-7y=11
2)x+y=7*5
5x-7y=11
3)5x+5y=35
5x-7y=11
4)(5x+5y)-(5x-7y)=35-11
5x+5y-5x+7y=24
5)12y=24
y=2
6)x+2=7
x=5
1)8x+3y=21
4x+5y=-7
2)8x+3y=21
4x+5y=-7*2
3)8x+3y=21
8x+10y=14
4)(8x+3y)-(8x+10y)=21-14
8x+3y-8x-10y=7
5)-7y=7
y=-1
6)4x-5=-7
x=0.5