SOLUTION: Elizabeth wishes to purchase $3 boxes of cookies and $5 boxes of cookies. She decides to buy 4 fewer $5 boxes than $3 boxes. How many of each box can she buy for no more than $20?

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Elizabeth wishes to purchase $3 boxes of cookies and $5 boxes of cookies. She decides to buy 4 fewer $5 boxes than $3 boxes. How many of each box can she buy for no more than $20?       Log On


   



Question 1173443: Elizabeth wishes to purchase $3 boxes of cookies and $5 boxes of cookies. She decides to buy 4 fewer $5 boxes than $3 boxes. How many of each box can she buy for no more than $20?

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
let x = the number of 3 dollars boxes.
let y = the number of 5 dollar boxes.

your equation will be 3x + 5y <= 20

the number of 5 dollars boxes is 1 less than the number of 3 dollars boxes.

the equation for that is y = x - 4

in the equation of 3x - 5y <= 20, replace y with x - 4 to get 3x + 5(x-4) <= 20

simplify this equation to get 3x + 5x - 20 <= 20

add 20 to both sides of this inequality and combine like terms to get 8x <= 40

solve for x to get x <= 5

since y = x - 4, then y <= 1

when x = 5, y = 1

3x + 5y <= 20 becomes 3*5 + 5*1 <= 20 which becomes 20 <= 20 which is true.

any value of x < 5 will result in y being equal to 0.

if she wants at least one 5 dollar box, then x must be equal to 5.