How can I find 2 equations with infinitely many solutions? Thanks

Answer 1

You have to make them multiplies of each other.

For example: x+ y = 4
2x + 2y = 8

If you try to solve these equations, both sides reduce to 0. 0 = 0 True statement!! This means that any x and y that work in one equation will work in the other equation.

Answer 2

yes, buying more ticetks increase the chance of winning.chance means percent probability, 40% chance means out of ten trials i'll have 4 winningdon't know what is your lottery scheme, in canada they have this lotto 649, there are 6 numbers , 1..2..3.. up to 49 for each number1st prize winning all six numbers could have a jackpot of $10 million or morethe lottery probability equation is as followsA=[ (a)(a-1)(a-2)(a-3)(a-4)(a-5) ]B=[ (b)(b-1)(b-2)(b-3)(b-4)(b-5) ]X=A / BP=1 /Xa=max for each number, for lotto 649, a=49b=numbers of number per ticket, for lotto 649, b=6P=probability in decimalexample for lotto 649A=[ (49)(49-1)(49-2)(49-3)(49-4)(49-5) ]B=[ (6)(6-1)(6-2)(6-3)(6-4)(6-5) ]X=[10 068 347 520 ] / [ 720 ]X=13 983 816the probability to win 6 number with one ticket is PP = 1 / XP = 1 / 13 983 816P = ( 7.151E-8 ) or ( 7.151E-6 ) %for two ticetksP = 2 / Xfor three ticetksP = 3 / X - etcif you are familiar with excel, you can use excel formula= HYPGEOMDIST ( 6, 6, 6, 49 )= 7.15112E-8= 7.15112E-6 %in lotto 649, winning three numbers out of 6 number, you will get a prize of $10, the chance of winning $10 is as follows= HYPGEOMDIST ( 3, 6, 6, 49 )= 0.017 650 404= 1.76 %not too bad although it cost $2 per ticket