Question

A student bought 1 box of crayons and 5 reams of paper for $54. She bought 5 boxes of crayons and 3 reams of paper for $50. What is the cost of each box of crayons and each ream of paper?

Answer 1

1 box of crayons costs $4.00
1 ream of paper costs $10.00

Explanation:

Two unknowns and two equations thus solvable.

Let 1 box of crayons cost be `c`
Let 1 ream of paper cost be `p`

condition(1): `->1c+5p=54" "` written as `c+5p=54`

condition(2): `->5c+3p=50`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

From condition(1)

Subtract `5p` from both sides.

`color(red)(c=54-5p)`

Substitute for `color(red)(c)` in condition(2) giving:

`color(green)(5color(red)(c)+3p=50 color(white)("ddd") ->color(white)("ddd")5(color(red)(54-5p))+3p=50)`

`color(green)(color(white)("ddddddddddddd")->color(white)("ddd")270-25pcolor(white)("d")+3p=50)`

`color(green)(color(white)("ddddddddddddd")->color(white)("ddd")270color(white)("ddd")-22pcolor(white)("dd")=50)`

Subtract 270 from both sides

`color(green)(color(white)("ddddddddddddd")->color(white)("dddd")0color(white)("dddd")-22p=-220`

Multiply both sides by (-1)

`color(green)(color(white)("ddddddddddddd")->color(white)("ddddddddddd")22p=220)`

divide both sides by 22

`color(green)(color(white)("ddddddddddddd")->color(white)("ddddddddddddd")p=10)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute for `color(red)(p=10)` in condition(1)

`color(green)(c+5color(red)(p)=54color(white)("dddd")->color(white)("dddddd")c+5(color(red)(10))=54)`

`color(green)(color(white)("ddddddddddddd")->color(white)("ddddd.d")c+color(white)("d")50color(white)("d")=54)`

Subtract 50 from both sides
`color(green)(color(white)("ddddddddddddd")->color(white)("ddddddd")c color(white)("d")=4)`