How do you solve by substitution # 8x - y = 4# and #5x + 3y = 17#?

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Answer 1 · 2015-06-20T12:38:44

Answer: #x=1,y=4#

#{(8x-y=4),(5x+3y=17) :}#

#{(y=8x-4),(5x+3y=17) :}#

Now we can substitute #8x+4# as #y# in the second equation:

#5x+3(8x-4)=17#
#5x+24x-12=17#
#29x=29#
#x=1#

Now we substitute #1# as #y# in #8x-y=4#

#8*1-y=4#
#y=4#

So we finally get: #{(x=1),(y=4):}#