Algebra b1 story problems

Question

W Auburn Locker selling wrapping papers for a school fundraiser customer can buy rolls of plain wrapping paper in rolls of shiny wrapping paper Wilbur sold three rolls of plain wrapping paper and 10 rolls of shiny wrapping paper for a total of 254 Mike told one roll of plain wrapping paper in 10 rolls of shiny wrapping paper for a total of 218 what is the cost of each one roll of plain wrapping paper on one roll of shiny wrapping paper

Answer 1

Assuming 254 and 218 are $2.54 and $2.28 (providing accurate information is important!)

1) 3p + 10s = 2.54

2) 1p + 10s = 2.18

An algebraic solution

Multiply equation 2) by 3

3) 3p + 30s = 6.54

Subtract 1) from 3)

20s = 4

s = 0.20

A logic solution:

2 additional p result in an additional 0.36

2p = 0.36

p = 0.18

Answer 2

There are two methods of solving this equation: Elimination and substitution.

Because both people sold 10 shiny rolls, the easiest method would be elimination.

Plain = p

Shiny = s

Wilbur sold 3 plain and 10 shiny for 254 so that would be

3p+10s = 254

Mike sold 1 plain and 10 shiny for 218 so that would be

1p+10s = 218

3p+10s = 254

1p+10s = 218

If we make one of the 10s negative they would cancel out. To do this you would have to distribute it to the whole equation so

-(1p+10s = 218)

becomes

-1p-10s = -218

now we have 3p+10s = 254

-1p-10s= -218

_____________

2 p = 36

p = 36/2

p = 18

Plain wrapping paper is 18

Now plug the 18 into one of the equations

3(18)+10s = 254

54 +10s = 254

10s = 254-54

10s=200

s = 200/10

s = 20

Shiny wrapping paper is 20

Plain wrapping paper is 18