
Carl M. answered 03/01/15
Tutor
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Pre-Algebra, Algebra 1, Algebra 2 and Geometry Tutoring
8r-s+t=-1 Eq. 1
2r+2s-3t=-22 Eq. 2
r-3s+2t=14 Eq. 3
t = -8r + s - 1 Solve Eq. 1 for t.
Substitute the value of t into Eq. 2.
2r 2s-3(-8r+s-1)= -22
2r+2s+24r-3s+3= -22
26r-s= -25
Substitute the value of t into Eq. 3.
r-3s+2(-8r+s-1)= 14
r-3s-16r+2s-2= 14
-15r-s= 16
Subtract Eq. 3 from Eq 2.
26r-s= -25 -
(-15r-s= 16)
41r= -41 This implies that r= -1.
26r-s= -25 Solve Eq. 2 for s by substituting the value of r.
26(-1)-s= -25
s= -1
t = -8r + s - 1 Solve Eq. 1 for t by substituting the values of r and s.
t = -8(-1) + (-1) -1
t = 8-1-1 = 6.
Check by substituting the values of r, s, and t into the original equations.
Eq. 1
8r-s+t=-1
8(-1)-(-1)+(6)=-1
-8+1+6=-1
-8+7=-1
-1=-1 Checks!
2r+2s-3t=-22
2(-1)+2(-1) -3(6)= -22
-2-2-18=-22
-22=-22 Checks!
r-3s+2t=14
-1-3(-1)+2(6)= 14
-1+3+12= 14
-1+15=14
14=14 Checks!