Решите системы уравнений способом подстановки:
3) (3x+y – 4 = 0,
5х + у — 10 = 0;
2) -x+y – 4 = 0,
— 4x+y+1 = 0;
4) (8x — Зу – 7 = 0,
13х +y – 9 = 0;
3) (3x+y – 4 = 0,
5х + у — 10 = 0;
2) -x+y – 4 = 0,
— 4x+y+1 = 0;
4) (8x — Зу – 7 = 0,
13х +y – 9 = 0;
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Ответ:
3) 3х+у-4=0
у=10-5х
3х+10-5х-4=0
-2х=-6
х=3
у=10-5×3=-5
2) у=4+х
4х+у+1=0
4х+4+х+1=5х+5=0
5х=-5
х=-1
у=4-1=3
4) 8х-3у-7=0
у=9-13х
8х-3(9-13х)-7=8х-27+39х=47х-27=0
47х=27
х=0,6
у=9-13×0,6= 1,2