document.write( "Question 535556: The concession stand is selling hot dogs and hamburgers during a game. At halftime, they sold a total of 78 hot dogs and hamburgers and brought in $105.50. How many of each item did they sell if hamburgers sold for $1.50 and hot dogs sold for $1.25? \n" ); document.write( "

Algebra.Com's Answer #351960 by KMST(5328) $\"\"$ $\"About$

You can put this solution on YOUR website!
IF YOU ARE STUDYING ALGEBRA,
\n" ); document.write( "you are used to working with variables, and know that the most popular names for those variables are x and y.
\n" ); document.write( "Let x be the number of burgers.
\n" ); document.write( "Let y be the number of hot dogs.
\n" ); document.write( "The amount, in dollars that hamburgers brought in is
\n" ); document.write( " $\"1.50%2Ax\"$ or $\"1.5%2Ax\"$
\n" ); document.write( "(Zeros at the end, after a decimal point have no meaning in algebra, and are ignored. Calculators ignore them too).
\n" ); document.write( "The amount, in dollars that hot dogs brought in is
\n" ); document.write( " $\"1.25%2Ay\"$
\n" ); document.write( "The total amount is
\n" ); document.write( " $\"1.5x%2B1.25y=105.5\"$
\n" ); document.write( "If decimals bother you (or your teacher), we can multiply everything by some number to get an equivalent equation without decimals:
\n" ); document.write( " $\"4%2A%281.5x%2B1.25y%29=4%2A105.5\"$ ---> $\"6x%2B5y=422\"$
\n" ); document.write( "The total number of items sold is
\n" ); document.write( " $\"x%2By=78\"$
\n" ); document.write( "Now we have a system of 2 linear equations with 2 variables to solve.
\n" ); document.write( "You probably know that there are two very popular methods to do it: substitution and elimination.
\n" ); document.write( "Substitution:
\n" ); document.write( "We solve for y (you could choose to solve for x):
\n" ); document.write( " $\"x%2By=78\"$ ---> $\"y=78-x\"$
\n" ); document.write( "We substitute that expression in the other equation:
\n" ); document.write( " $\"6x%2B5%2A%2878-x%29=422\"$ ---> $\"6x%2B5%2A78-5%2Ax%29=422\"$ ---> $\"6x%2B390-5x%29=422\"$ ---> $\"x%2B390=422\"$ ---> $\"x=422-390\"$ ---> $\"x=32\"$
\n" ); document.write( "Then we substitute that value for x in $\"y=78-x\"$
\n" ); document.write( " $\"y=78-32\"$ ---> $\"y=46\"$
\n" ); document.write( "IF YOU NEVER STUDIED ALGEBRA your strategies would be guess and check or you could calculate the effect of talking each person wanting a hot dog into ordering a hamburger instead. (For only an extra $0.25, they get more protein and less fat). If everyone ordered hot dogs, all 78 items sold would be hot dogs and the total would be $97.50. To get $105.50 you need another $8.00, meaning 32 of those orders to switch from hot dog to burger, leaving 32 burgers and 78-32=46 hot dogs.
\n" ); document.write( "Guess and check would start by guessing a number for hot dogs of hamburgers and checking the number for the other item and the total.
\n" ); document.write( "You could guess 40 hot dogs ($50=40*$1.25), so that would mean 38 (78-40=38) hamburgers ($57=38*$1.50) That would be a total of $107, which is too much. Less of the expensive item (the burger) was sold. So you try 50 hot dogs ($62.50=50*$1.25), which leaves 28 (78-50=28) burgers ($42=$1.50*28), for a total of $104.50, which is too little, but you are a bit closer. You need a guess between 40 and 50 hot dogs, a bit closer to 50 than to 40. Guessing 46 hot dogs would get you the answer. Other wise, you would need one or two more guesses.
\n" ); document.write( " \n" ); document.write( "

\n" );