How do you solve 3r-5s=-35 and 2r-5s=-30?

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Answer 1 · 2015-09-05T07:19:51

The solution is:
#color(blue)(r=-5, s=4#

#3r-color(blue)(5s)=-35# ....equation #1#
#2r-color(blue)(5s)=-30#......equation #2#

We can solve the system of equations by elimination as subtracting the two would eliminate #color(blue)(5s#

Subtracting equation #2# from #1#
#3r-cancelcolor(blue)(5s)=-35#
#-2r+cancelcolor(blue)(5s)=+30#

#color(blue)(r=-5#

Finding #s# by substituting the value of #r# in equation #1#
#3r-5s=-35#

#3r+ 35=5s#

#3 * (color(blue)(-5))+ 35=5s#

#-15+ 35=5s#

#20 = 5s#

#20/5 = s#

#color(blue)(s=4#