How do you solve the system #3x+2y=-17# and #2x-5y=14# by substitution?

2 Answers
EZ as pi
Mar 10, 2018

#x =-3 and y =-4#

Explanation:

Substitution would certainly not be my method of choice for this example, but here goes...

You need to have one of the equations with #x# or #y# as the subject, ie as a single variable,

#2x -5y = 14# can be written as # color(blue)(x= (5y+14)/2)#

Now substitute that expression for #x# in the other equation:

#3color(blue)(x)+2y=-17#

#3color(blue)(((5y+14)/2)) +2y =-17: "larr# now solve for #y#

#3(5y+14)+4y=-34" "larr xx2#

#15y+42+4y=-34#

#19y = -34-42#

#19y = -76#

#y=-4#

Now find #x# by substituting #y=-4#

# color(blue)(x= (5(-4)+14)/2)#

#x =-3#

Check:

#3x+2y#

#=3(-3)+2(-4)#

#-9-8#

# = -17#

Barney V.
Mar 10, 2018

#y=-4,x=-3#

Explanation:

#3x+2y=-17------(1)#

#2x-5y=14-------(2)#

#(1)xx2:-#

#:.6x+4y=-34------(3)#

#(2)xx3:-#

#:.6x-15y=42------(4)#

#(3)-(4):-#

#:.19y=-76#

#:.y=-4#

substitute # y=-4 # in (1)#

#:.3x+2(-4)=-17#

#:.3x-8=-17#

#:.3x=-17+8#

#:.3x=-9#

#:.x=-9/3#

#:.x=-3#
~~~~~~~~~~~~~~
substitute #y=-4,x=-3# in (2)

#:.2(-3)-5(-4)=14#

#:.-6+20=14#

#:.14=14#

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