How do solve the following linear system?: # 4x+3y=1 , 11x + 3y + 7 = 0 #?

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Answer 1 · 2017-05-05T06:37:35

#x=-8/7=-1 1/7#

#y=13/7=1 6/7#

Given: #4x+3y=1#; #11x+3y+7=0#

From #4x+3y=1# we can subtract #4x# from both sides

to get #3y=1-4x# which we can substitute into the other

equation for #3y#:

#11x+3y+7=0#

#11x+1-4x+7=0#

#7x=-8#

#x=-8/7# answer x

Substitute value for #x# into #4x+3y=1# to find #y#:

#4(-8/7)+3y=1#

#3y=1+32/7#

#3y=39/7#

#y=13/7# answer y

To check, substitute into the #given# equation:

#11x+3y+7=0#

#11(-8/7)+3(13/7)+7=0#

#(-88/7)+(39/7)+(49/7)=0#

#(-88)+(39)+(49)=0#

#-88+88=0#