SOLUTION: c/2-d/5=-4 and c-d/5=-8 solve by substitution method

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Question 126976This question is from textbook algebra structure&method book 1
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c/2-d/5=-4 and c-d/5=-8 solve by substitution method This question is from textbook algebra structure&method book 1

Found 2 solutions by josmiceli, checkley71:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
I notice that d%2F5 is common to both
c%2F2+-+d%2F5+=+-4
c+-+d%2F5+=+-8
add d%2F5 to both sides
c+=+-8+%2B+%28d%2F5%29
add 8 to both sides
d%2F5+=+c+%2B+8
substitute this into %28c%2F2%29+-+%28d%2F5%29+=+-4
%28c%2F2%29+-+%28c+%2B+8%29+=+-4
%28c%2F2%29+-+c+-+8+=+-4
-%28c%2F2%29+=+4
c+=+-8 answer
d%2F5+=+c+%2B+8
d%2F5+=+-8+%2B+8
d+=+0 answer

Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
C-D/5=-8 OR C=D/5-8 OR C=(D-40)/5
[(D-40)/5)]/2-D/5=-4
(D-40)/5*1/2-D/5=-4
(D-40)/10-D/5=-4
(D-10-2D)/10=-4
(-D-10)/10=-4
-D-10=-40
-D=-40+10
-D=-30
D=30 ANSWER.
C-(30/5)=-8
C-6=-8
C=-8+6
C=-2 ANSWER.
PROOF:
-2-30/5=-8
-2-6=-8
-8=-8