Answer:
The solution is x=1,y=2,z=3
Step-by-step explanation:
The given system of equations is ;\
2x−3y+4z=8...(1)
3x+4y−5z=−4...(2)
4x−5y+6z=12...(3)
Make x the subject in equation (1)
Put equation (4) into equation (2) and (3)
Multiply through by;
Expand;
Simplify;
Equation (4) in (3)
Put equation (6) into equation (5)
z=3
Put z=3 into equation (6)
y=2(3)-4=2
Put y=2 and z=3 into equation 4
The solution is x=1,y=2,z=3
There are 45 numbers out of which 5 numbers are required to win an enormous sum of money
No of ways in which 5 numbers can be selected out of 45 numbers.
out of which there is only combination foe contest winning
When 100 different tickets are bought then
Probability of winning:
Probably of winning≈0,000082
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P.S. Hello from Russia :^)
