How do you solve #5x + 8y = 7# and #x + y = 2# using substitution?

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Answer 1 · 2016-04-24T13:22:42

x=3, y= -1

given x+y=2
y= 2 - x (eq. 1)

given 5x+8y=7 (eq.2)

substitute eq. 1 in eq. 2

5x + 8(2-x) =7
5x + 16 - 8x = 7 (open the bracket)
-3x=7-16= -9
x= -9/ -3 = 3

eq 1-
y= 2-3= -1

so x=3, y= -1

Answer 2 · 2016-04-24T13:23:00

#(x,y)=(3,-1)#
See below for method of solution by substitution.

If #x+y=2#
then #y=2-x#
and
#color(white)("XXX")(2-x)# can be substituted for #y# in #5x+8y=7#

This would give:
#color(white)("XXX")5x+8(2-x)=7#

#color(white)("XXX")5x+16-8x=7#

#color(white)("XXX")-3x=-9#

#color(white)("XXX")x= 3#

Then, returning to #y=2-x#
we have
#color(white)("XXX")y=2-3 =-1#