#"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#