Math, asked by kav1, 1 year ago

Solve x+y+z=100 and 40x+1/3y+1/5z=100

Answers

Answered by vinosharubalitanu

2

[40x+x]+[3y+y]+[5z+5z]+1+1=100+100
[40*2x]+[3*2y]+[5*2z]+2=200
80x+6y+10z=200-2
80x+6y+10z=198

Answered by nabiha55

0

i also need this answer

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