Here are the equations. I need to solve for D. 2B + 12.67D = 8350 --------------

Question

Here are the equations. I need to solve for D.

2B + 12.67D = 8350   ------------------------(1)

10.67 D – 2A = 6350  -----------------------(2)

8.35A + 6.35B – 4.32D = 0  ----------------(3)

12.67A + 10.67B = 4320   ------------------(4)

My answer:

From (2), A = (10.67D – 6350)/2

From (1), B = (8350 – 12.67D)/2

Substituting to equation (3), 8.35[(10.67D –6350)/2] + 6.35[(8350 – 12.67D)/2] – 4.32D = 0  

(44.54725D – 26511.25) + (26511.25 –40.22725D) – 4.32D = 0

D = 0

Thus, if I substitute A and B to equation (3), D is zero.

If I substitute A and B to equation (4), D is also zero.

However, another method makes D not zero. That's why I said it'sstrange.

From (2), D = (6350+ 2A)/10.67

Substituting to equation (3),

8.35A + 6.35B – 4.32[(6350+ 2A)/10.67] =0   

7.54A + 6.35B = 2570.95 -----------------------------(5)   

12.67A + 10.67A = 4320   -----------------------from(4)

I can solve for A and B from equation (4) and (5) usingcalculator. A =-13.5, B = 420.93

From equation (1), 2(420.93) + 12.67D = 8350  

Explanation / Answer

assume that what you have is correct for D where: D = 0  then applying that into equation (1) and solve for B you get: B =4175 similarly applying it to equation (2) and solve for A you get: A =-3175 use either equation (3) or (4) to check you answer i.e.    equation (3): 8.35(-3175) + 6.35(4175) = 0      => 0 = 0        so thatcheck similarly       equation(4):      12.67(-3175) + 10.67(4175) =4320                                                                                     4320 = 4320        so that checkas well A = -3175, B = 4175, and D = 0

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