0.5(2y-1)-(0.5-0.2y)+1=0 и 0,3(3x-1)+0.2=5(0.1-0.2x)-0.1 срочно

Question

0.5(2y-1)-(0.5-0.2y)+1=0 и 0,3(3x-1)+0.2=5(0.1-0.2x)-0.1 срочно

Answer 1

1)1y-0.5-0.5+0.2y+1=0
1.2y-1+1=0
1.2y=0
y=0
2)0.9x-0.3+0.2=0.5-x-0.1
0.9x-0.1=0.4-x
0.9x+x=0.4+0.1
1.9x=0.5
x=0.5/1.9
x=2.6

Comments

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