algebra word problem

Question

tyler is catering a banquet for 250 people. each person will be served either a chicken dish that cost $5 eachor a beef dish that cost $7 each. Tyler spent $1500. how may dishes of each type did tyler serve

Answer 1

Dear Norma,

 

This type of problem is called "Systems of Equations". Basically you have two unknowns, people who ordered chicken, and people who ordered beef. Luckily, you have two INDEPENDENT pieces of information, so you can make two equations.

 

Let x = chicken meals,

y = beef meals.

 

First equation: X+Y = 250  --- total people at the banquet

Second eq: 5*X + 7*Y = $1500 --  the total price of the food is the SUM of chicken @ 5$ per piece, and beef $ 7$ per piece

 

so now, you need to find a way to eliminate one variable in one of the equations. Lets eliminate X in equation 1:

Multiply the whole equation 1 by -5:

 

-5X - 5Y = -1250

 

now ADD the left side of 1 to left side of #2, and the right side of eq 1 to right side of eq 2:

-5X+5X - 5Y + 7Y = 1500 - 1250 

simplified becomes

2y=250, or 

 

y=125

 

now go back to equation 1  and plug in for y:

 

X+125 = 250

 

X = 125

 

So its 125 meals of each kind, chicken and beef.

 

Khan Academy has a great lesson on solving systems of equations! You should check it out.

Cheers,

kyle

 

Answer 2

let x= chicken and y= beef

x+y=250 people also

5x+7y=1500 cost

solve the two equations

we will multiply the first one by -5

-5x-5y=-1250

5x+7y=1500

so the x cancel and end up with 2y=250

divide by 2 y=125 will eat beef then

125 will also eat chicken