Question

Marcus purchased 5 notebooks and 10 boxes of crayons for $31. Nina went to the same store and bought 10 notebooks and 5 boxes of crayons for $24.50. What is the cost of one notebook and one box of crayons?

Answer 1

`x=1.20`
`y=2.50`

Explanation:

`"Solving Process:"`
Let:
`x= "the price of the notebooks"`
`y= "the price of the boxes of crayons"`

Now, formulate equations with reference to their purchases; that is,
`color(red)("Marcus ": 5x+10y=31->eq.1`
`color(blue)("Nina ": 10x+5y=24.50->eq.2`

Then, solve the equations simultaneously as follows:

Multiply eq.1 with 2 to eliminate the terms with x variable in both equations.

`eq.1-> color(red)(5x+10y=31) }-2`
`eq.2->color(blue)(10x+5y=24.5`

`"so that the eq.1 becomes"`

`eq.1->color(red)(cancel(-10x)-20y=-64`
`eq.2->color(blue)(cancel(10x)+5y=24.5`

Then find the difference of the remaining terms to get the equation as shown below and find the value of `y`.

`color(red)(-15y=-37.5)`; divide both sides by `-15` to isolate `y`
`color(red)((cancel(-15)y)/(cancel(-15))=(-37.5)/(-15))`
`color(red)(y=2.50`; price for the boxes of crayons

Now, find the value of `x`, the price of the notebooks, by using either of the equations formulated. Here, eq.1 is used to solve for `x`.

`color(red)(5x+10y=31)`; where `color(red)(y=2.50)`
`color(red)(5x+10(2.50)=31)`; simplify
`color(red)(5x+25=31)`; combine like terms
`color(red)(5x=31-25)`; simplify
`color(red)(5x=6)`; isolate `x` by dividing both sides by `5`
`color(red)(x=1.20)`; the price of the boxes of crayons

`"Checking Process":`
where: `x=1.20 and y=2.50`

`Eq.1`
`5x+10y=31`
`5(1.20)+10(2.50)=31`
`6+25=31`
`31=31`

`Eq.2`
`10x+5y=24.5`
`10(1.20)+5(2.50)=24.5`
`12+12.5=24.5`
`24.5=24.5`