How do you solve the following system?: # y = 3/4x + 5 , x + 2y = 3 #

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Answer 1 · 2016-02-27T08:58:05

#x=-14/5# and #y=29/10#

From the given equations
#y=(3x)/4+5# and #x+2y=3#

Using the second equation #x+2y=3#
Let #x=3-2y# then substitute in the first equation

first equation becomes

#y=(3(3-2y))/4+5#

multiply both sides by of the equation by 4 now to get rid of any fraction

#4y=4*(3(3-2y))/4+4*5#

#4y=cancel4*(3(3-2y))/cancel4+4*5#

#4y=9-6y+20#

#4y+6y=9+20#

#10y=29#
divide both sides by 10

#(10y)/10=29/10#

#(cancel10y)/cancel10=29/10#

#y=29/10#

Solve now for the value of x using the second equation

#x=3-2y# and #y=29/10#

#x=3-2*(29/10)#
#x=3-29/5#
#x=(15-29)/5#

#x=-14/5#

God bless...I hope the explanation is useful..