How do you solve the following system using substitution?: #3s-5t=-30, 7s+11t=32#

1 Answer
Alan P.
Oct 23, 2015

#(s,t)=(-5/2,9/2) = (-2 1/2, 4 1/2)#

Explanation:

Given
[1]#color(white)("XXX")3s-5t=-30#
[2]#color(white)("XXX")7s+11t=32#

Rewriting [1] to generate an equation of the form: #s = # some expression
[3]#color(white)("XXX")3s = 5t-30#

[4]#color(white)("XXX")s = (5t-30)/3#

We can now substitute (as per the question) #((5t-30)/3)# for #s# in [2]
[5]#color(white)("XXX")7((5t-30)/3)+11t = 32#

Simplifying
[6]#color(white)("XXX")(35t-210 +33t)/3 = 32#

[7]#color(white)("XXX")(68t-210) = 96#

[8]#color(white)("XXX")68t = 306#

[9]#color(white)("XXX")t = 9/2#

Substituting #9/2# into [1] in place of #t#
[10]#color(white)("XXX")3s -5*(9/2) = -30#

[11]#color(white)("XXX")3s = (5*9 -2*30)/2#

[12]#color(white)("XXX")3s =-15/2#

[13]#color(white)("XXX")s = - 5/2#

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