SOLUTION: Solve by the substitution method 3d+e over 4 = d+1 over 2 d-e over 4 = 1 If anyone can answer this problem I would greatly appreciate it. Please and Thank You

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Question 119577This question is from textbook Algebra Structure and Method
: Solve by the substitution method
3d+e over 4 = d+1 over 2
d-e over 4 = 1


If anyone can answer this problem I would greatly appreciate it. Please and Thank You This question is from textbook Algebra Structure and Method

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Solve by the substitution method
3d+e over 4 = d+1 over 2
d-e over 4 = 1
;
%28%283d%2Be%29%29%2F4 = %28%28d%2B1%29%29%2F2
Cross multiply and you have:
2(3d+e) 4(d+1)
6d + 2e = 4d + 4
6d - 4d + 2e = 4
2d + 2e = 4
Simplify, divide by 2:
d + e = 2
e = (2-d); Use this for substitution
:
%28%28d-e%29%29%2F4 = 1
Multiply equation by 4 to get rid of that annoying denominator, we have:
d - e = 4
:
Substitute (2-d) for e in the above equation:
d - (2 - d) = 4
d - 2 + d = 4
2d = 4 + 2
2d = 6
d = 3
:
Find e substituting 3 for d in equations; e = (2 - d)
e = 2 - 3
e = -1
:
:
Check solutions in the original 1st equation:
%28%283d%2Be%29%29%2F4 = %28%28d%2B1%29%29%2F2
%28%283%283%29-1%29%29%2F4 = %28%283%2B1%29%29%2F2
%28%288%29%29%2F4 = %28%284%29%29%2F2; confirms our solutions
:
You can check solution in the original 2nd equation