How do you solve #2p + q = 1# and #9p + 3q + 3 = 0 # using substitution?

1 Answer
Alan P.
Apr 8, 2016

#(p,q)=(-2,5)#
(see below for substitution methodology)

Explanation:

Given #2p+q =1#
#rarr color(red)(q)=(color(blue)(1-2p))#

Also given #9color(green)p+3color(brown)(q)+3=0#

Substituting #(color(blue)(1-2p))# for #color(brown)(q)#

#color(white)("XXX")9color(green)(p)+3(color(blue)(1-2p))+3=0#

#color(white)("XXX")3p+3+3=0#

#color(white)("XXX")color(cyan)p=(color(orange)(-2))#

Substituting #(color(orange)(-2))# for #p# in the original equation:
#color(white)("XXX")2(color(orange)(-2))+q=1#

#color(white)("XXX")q=5#

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