Let #(x)/(y)# represent the fraction
The problem specifies
#(a)# Adding #1# to the numerator and denominator makes the fraction equal to #(1)/(2)#
#(b)# Subtracting #1# from the top and bottom makes the fraction equal to #(1)/(3)#
#(a) (x + 1)/(y+1) = (1)/(2)#
#(b) (x - 1)/(y-1) = (1)/(3)#
#color(white)(........................)#―――――――
Solve for #x#, already defined as "the numerator of the original equation"
These equations can be solved as Ratio&Proportion problems.
#1)# Cross multiply both equations
#(a)# #2(x+1)=1(y+1)#
#(b)# #3(x−1)=1(y−1)#
#2)# Clear the parentheses
#(a)# #2x + 2 = y + 1#
#(b)# #3x - 3 = y - 1#
#3)# Write the equations with the variables on the left and the numbers on the right
#(a)# #2x - y =# - #1#
#(b)# #3x - y = 2#
#4)# Start with Equation #(b)# and subtract Equation #(a)#
(Do it in this order to avoid negative numbers.)
#(b)# #3x - y = 2#
#(a)# - (#2x - y =# - #1#)
#5)# Clear the parentheses and combine to eliminate the #y# term
#(b)# #3x - y = 2#
#(a)# #-2x + y = 1#
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#color(white)(..........) x# #color(white)(...m)= 3# #larr# already defined as "the numerator of the original fraction"
#color(white)(........................)#―――――――
Solve for #y#, already defined as "the denominator of the original fraction"
#1)# Using one of the original equations, sub in #3# in the place of #x#
#(color(blue)(x) - 1)/(y-1) = (1)/(3)#
#(color(blue)(3) - 1)/(y-1) = (1)/(3)#
#2)# Cross multiply and solve for #y#, already defined as "the denominator of the original fraction"
#3(3-1)=1(y-1)#
#3)# Solve inside the parentheses
#3(2)=1(y-1)#
#4)# Clear the parentheses by distributing the #3# and the #1#
#6 = y - 1#
#5)# Add #1# to both sides to isolate #y#
#6)# #7 = y# #larr# answer for #y#, defined as "the denominator of the original fraction"
#"Answer"#
The original fraction is #(3)/(7)#
#color(white)(........................)#―――――――
Check
Add #1# to the top and bottom of #(3)/(7)# to see if it becomes #(1)/(2)#
#(3 + 1)/(7+1) =# #(4)/(8) = (1)/(2)# ✓
Subtract #1# from the top and bottom to see if it becomes #(1)/(3)#
#(3-1)/(7-1) =# #(2)/(6) = (1)/(3)# ✓
#Check#
