#"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#