How do you solve the simultaneous equations #3x + 5y = 11# and #2x + y = 3#?

2 Answers
Gió
Jul 22, 2015

I found:
#x=4/7#
#y=13/7#

Explanation:

From the second equation you can isolate #y# as:
#y=3-2x#
substitute nto the first to find #x#:
#3x+5(color(red)(3-2x))=11#
#3x+15-10x=11#
#-7x=-4#
#x=4/7#
substitute back into the second equation:
#y=3-2(4/7)=(21-8)/7=13/7#

Alan P.
Jul 22, 2015

#(x,y) = (4/7, 13/7)#

Explanation:

[1]#color(white)("XXXX")##3x+5y = 11#
[2]#color(white)("XXXX")##2x+y = 3#

subtract #2x# from both sides of [2] to isolate #y# on the left side
[3]#color(white)("XXXX")##y = 3-2x#

substitute #(3-2x)# for #y# in [1]
[4]#color(white)("XXXX")##3x + 5(3-2x) = 11#

simplify
[5]#color(white)("XXXX")##-7x + 15 = 11#

[6]#color(white)("XXXX")##x = 4/7#

substituting #4/7# for #x# in [2]
[7]#color(white)("XXXX")##2(4/7) + y = 3#

[8]#color(white)("XXXX")##y = 13/7#

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