How do you solve the system #x/2+(2y)/3=32# and #x/4-(5y)/9=40#?

1 Answer
Feb 22, 2017

Answer:

#x = 100 and y = -27#

Explanation:

At first glance this horrifying because of the fractions!

Luckily with equations you can always get rid of any fractions by multiplying each term by the LCM of the denominators.

#xx6 rarr " "x/2+(2y)/3=32" "rarr 3x+4y = 192#

#xx 36rarr" "x/4-(5y)/9=40" "rarr 9x -20y = 1440#

The equations look much better, now we can solve them:

#color(white)(..............)3x+4y = 192......................................A#
#color(white)(..............)9x-20y= 1440...................................B#

#A xx-3:" "color(blue)(-9x)-12y = -576......................C#
#color(white)(........................)color(blue)(9x)-20y= 1440.........................B#

#C+B:" "-32y =864#
#color(white)(................................)color(red)(y=-27)#

Substitute #-27 # for #y# in #A#

#" "3x+4(-27) = 192......................................A#
#" "3x-108 = 192#
#" "3x = 192+108#
#" "3x = 300#
#" "color(blue)(x = 100)#

Check the solutions in equation B

#" Is " 9(100)-20(-27)= 1440 ?#
#" "900+540= 1440#
#" Indeed! "1440 = 1440#

We can even check in the original equation for A;

#x/2+(2y)/3=32#

#100/2 +(2xx-27)/3#

#=50-18#

#=32" "larr# the answer is correct.