How do solve the following linear system?: #3s + 4 = -4t, 7s + 6t + 11 = 0 #?

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Answer 1 · 2018-07-05T14:35:49

#s=-2# and #t=1/2#

From first equation, #s=(-4t-4)/3#.

After plugging value of #s# into second one,

#7*(-4t-4)/3+6t+11=0#

#(-28t-28+18t+33)/3=0#

#(5-10t)/3=0#

#5-10t=0#

#10t=5#, so #t=5/10=1/2#

Thus, #s=((-4)*1/2-4)/3=-2#

Answer 2 · 2018-07-05T14:37:57

Your answer is #-28/10=t #

Here we have two equation
1) #3s + 4 = - 4t#
2) #7s + 6t = 0#

using equation 2 to find value of s

#7s+ 6t = 0#
#7s=-6t#
#s=-(6t)/7# ....(equation3)

Now put value of s in eq 1

#3((-6t)/7) + 4 =-4t#

#(-18t+28)/7=-4t#

#-18t+28=-28t#

# 28= -28t+18t#

#28=-10t#

#-28/10=t #

Now put value of t in equation 3

#s=-(6(-28/10))/7#

#s=6(28/70)#