What is the solution set of the equation #x/5 + x/2 =14#?

Question

Ethan Adkins · Accepted Answer

x=20

#(2x)/10 + (5x)/10 = (7x)/10 = 14# #7x = 140# #x = 140/7# #x=20#

Kennedy Adams · Answer

Virtually every step explained. Once practised you will start to use shortcut methods and jump steps (much faster).

#x=20# Fraction structure: #ubrace(color(purple)(("count")/("size indicator of what is being counted") ))color(white)("d")=color(white)("d")ubrace(color(red)( ("numerator")/("denominator"))) # #color(white)("ddd")ul("My")" description of purpose"color(white)("dddddddddd") "The proper names" # #color(white)("dddddddddddddddddddddddddddddddddd")"that people use"# You can only DIRECTLY add or subtract the top numbers (counts) unless the bottom numbers (size indicators) are the same. Multiply by 1 and you do not change the value. However 1 comes in many forms. #color(green)([x/5color(red)(xx1)]+[ x/2color(red)(xx1)] = 14 )# #color(green)([x/5color(red)(xx2/2)]+[ x/2color(red)(xx5/5)] = 14 )# #color(green)(color(white)("d")[(2x)/10]color(white)("dd")+color(white)("dd")[ (5x)/10] color(white)("d")= 14 )# As the bottom numbers are now the same we can do this: #color(green)((2x+5x)/10 = 14# #color(green)((7x)/10 = 14# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(brown)("Now we determine the value of "x)# Multiply both sides by 10 #10/10xx7x=14xx10# but #10/10=1# giving: #7x=140# Divide each side by 7 #7/7xx x=140/7# We can 'split' 140 into #14xx10# #x=(14xx10)/7# #x=14/7xx10# #x=2xx10=20#