Solving voter fraud with algebra!

[Billo_Really]

I think it's safe to say, if you are an American, born and raised in this country, you've had to grind through algebraic word problems at some point in middle school. So, I propose, in order to ensure only American's are voting in the polls, we put a word problem at the top of registration cards. This should maintain the purity of our elections.

Sample algebra word problem:
A dealer has some hard candy worth $2.00 a pound and some worth $3.00 a pound.
He wishes to make a mixture of 80 pounds that he can sell for $2.20 a pound.
How many pounds of each candy does he need?

Answer:
Let amount of $2 hard candy be "x".
Then amount of $3 hard candy is "80-x"
Value of $2 candy is 2x dollars
Value of $3 candy is 3(80-x)=240-3x dollars
Value of the mixed candy is 2.2(80)=176 dollars

Equation:
Value of $2 candy + Value of $3 candy = Value of mixed candy
2x + 240-3x=176
-x=-64
x= 64 pounds of $2 hard candy
80-x=16 pounds of $3 hard candy.​

Granted, some bible belt crackers may have some difficulty with the test, but the homies from South Central won't, thus ensuring the purity of the election.

 

[JustTheFacts]

Pfft

The answer is simple. In this country you have to have a state ID to fly, to rent a hotel, to rent a car, to drive a car, to buy a beer, to buy an r rated movie, to buy tobacco, but not to vote?

 

[mjollnir]

Pfft

The answer is simple. In this country you have to have a state ID to fly, to rent a hotel, to rent a car, to drive a car, to buy a beer, to buy an r rated movie, to buy tobacco, but not to vote?

Voting is a right. What you listed are privileges.

Glad I could clear that up for you.

 

[1Templar]

E=mc2