How do you solve this system of equations: #4s - 3t = 8 and t = - 2s - 1#?

2 Answers
Alan P.
Oct 23, 2017

#(s,t)=(1/2,-2)#

Explanation:

Given
[1]#color(white)("XXX")4s-3t=8#
[2]#color(white)("XXX")t=-2s-1#

Based on [2] we can substitute #(-2s-1)# for #t# in [1]
[3]#color(white)("XXX")4s-3(-2s-1)=8#

Simplifying
[4]#color(white)("XXX")4s +6s+3=8#

[5]#color(white)("XXX")10s=5#

[6]#color(white)("XXX")s=1/2#

Substituting #1/2# for #s# in [2]
[7]#color(white)("XXX")t=-2 * (1/2) -1=-2#

Manasvini Nittala
Oct 23, 2017

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

Explanation:

the given two equations are
#4s-3t=8#
#t=-2s-1rArrt+2s=-1#
multiplying #2s+t=-1# by 3 we get
#3(2s+t=-1)rArr6s+3t=-3#
now on adding #(4s-3t=8)+(4s+3t=-3)#
we get #2s=1rArrs=1/2#
on substituting #s=1/2# in #4s-3t=8rArr4(1/2)-8=3trArr3t=-6rArrt=-2#
hence we get the answer #s=1/2 and t=-2#

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