The sum of two numbers is 12 Their difference is 4. How do you find the numbers?

Let the 2 numbers be x and y with x > y

#rArrxcolor(red)(+y)=12to(1)larr"sum of 2 numbers"#

#rArrxcolor(red)(-y)=4to(2)larr" difference of numbers"#

Adding the 2 equations, term by term on both sides, will eliminate y leaving an equation in x that we can solve.

#rArr(1)+(2)" gives"#

#(x+x)+(color(red)(-y+y))=(4+12)#

#rArr2x=16#

divide both sides by 2

#(cancel(2) x)/cancel(2)=16/2#

#rArrx=8#

Substitute this value into equation ( 1 ) and solve for y

#rArr8+y=12#

#rArry=12-8=4#

#"Thus the 2 numbers are " 8" and " 4#

let these two numbers be #a# and #b#.

#a+b=12#
#a-b=4#

if you add the two equations:

#a+b=12#
#a-b=4#

#2a+0 = 8#
#2a = 8#

divide by #2#:
#a = 4#

substitute:
#a+b = 12#
#4+b=12#
#b=12-4#
#=8#

#a=4#
#b=8#

#8+4=12#
#8-4=4#

Let #x# equal one number and #y# the other number.

One equation would be the sum of the numbers.

#x + y = 12#

the other equation would be the difference of the two numbers

#x-y = 4" "#

Now add the two equations

#x +y = 12 " " and " "x -y = 4" "#(adding #+ y and -y = 0#)

so #2x = 16" "# Now divide both sides by 2

# (2x)/2 = 16/2" "# This gives

#x = 8# Now substitute x into one of the equations and solve for y

#8 + y =12" "# subtract 8 from both sides

# 8 -8 +y = 12 -8 " "# This gives

#y = 4" "# Put values into the second equation to check

#8 -4 = 4 #

#4 = 4" "# check

#x = 8 and y = 4#

Questions which involve two or more numbers can be done using one variable to define them all.

Let the smaller number be #color(red)(x)#.

The difference between the numbers is #4#.

The other number is #color(blue)(x+4)#

Their sum is #12#

#color(red)(x) +color(blue)(x+4) =12#

#2x = 12-4#

#2x = 8#

#x =4#

#:. x+4 = 8#

The numbers are #4 and 8#