用加减消元法解下列方程组①x+y=36,x+2y=50 ②5x-6y=1,2x-6y=10

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用加减消元法解下列方程组①x+y=36,x+2y=50 ②5x-6y=1,2x-6y=10
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用加减消元法解下列方程组①x+y=36,x+2y=50 ②5x-6y=1,2x-6y=10
用加减消元法解下列方程组①x+y=36,x+2y=50 ②5x-6y=1,2x-6y=10

用加减消元法解下列方程组①x+y=36,x+2y=50 ②5x-6y=1,2x-6y=10
x+y=36,①
x+2y=50 ②
②-①得:
y=14
①×2-②得:
x=22
∴x=22
y=14
5x-6y=1,①
2x-6y=10 ②
①-②得:
3x=-9
x=-3
①×2-②×5得:
-12y+30y=2-50
18y=-48
y=-8/3
∴x=-3
y=-8/3

1.2y-y=50-36 y=14 x=22
2.5x-2x=1-10 x=-3 y=-8/3

x+y=36①
x+2y=50 ②
②-①:y=14,则x=22
5x-6y=1①
2x-6y=10②
②-①:-3x=9,x=-3,则y=-8\3