How do you find two numbers whose difference is 4 and the difference of their square is 88?

1 Answer

#x=13#
#y=9#

Explanation:

How do you find two numbers whose difference is 4 and the difference of their square is 88?

Solution:

let #x=#the larger number
let #y=#the smaller number

#x-y=4#first equation
#x^2-y^2=88#second equation

Substitute the first in to the second equation

#x^2-y^2=88#second equation
#(y+4)^2-y^2=88#
#y^2+8y+16-y^2=88#
#8y=88-16#
#8y=72#
#y=9#

Solve for #x# using first equation

#x-y=4#first equation
#x-9=4#
#x=9+4#
#x=13#

God bless... I hope the explanation is useful.