How do you solve 3/4x + 1/3y = 1 and x - y = 10?

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Answer 1 · 2018-07-14T11:52:50

#x=10+y# substitute this into the first equation

#3/4(10+y)+1/3y=1#

#30/4+3/4y+1/3y=1#

Multiply by 12 to remove the fractions

#90+9y+4y=12#

#90+13y=12#

#13y=-78#

#y=-6#

Put #y=-6 into equation 2

#x- -6=10#

#x=4#

Answer 2 · 2018-07-14T14:29:47

#x=4,y=-6#

#:.3/4x+1/3y=1---(1)#

#:.x-y=10----(2)#

#:.(2) xx 3/4#

#:.3/4x-3/4y=15/2----(3)#

#:.(1)-(3)#

#:.13/12y=-13/2#

#:.y=-13/2-:13/12#

#:.y=-cancel 13^1/cancel2^1 xx cancel12^6/cancel 13^1#

#:.y=-6#

#"substitute" # y=-6#" in" (2)#

#:.x-(-6)=10#

#:.x+6=10#

#:.x=10-6#

#:.x=4#

~~~~~~~~

check:-

#"substitute"# y=-6# and #x=3 #"in"(1)#

#:.3/4(4)+1/3(-6)=1#

#:.3-2=1#

#:.1=1#