Question 578380: Points A, B, C, and D are collinear. AB is 15 more than BC, and CD is 8 more than BC. If AD is 5 less than twice AB, what are the lengths of AB, BC, and CD? (hint: draw a diagram and use a variable to write an equation then solve)
I am having trouble figuring out the answer the this problem, I don't know how to get an equation to solve the lengths using the information I am given. Could you please help? Thanks
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Points A, B, C, and D are collinear. AB is 15 more than BC, and CD is 8 more than BC. If AD is 5 less than twice AB, what are the lengths of AB, BC, and CD? (hint: draw a diagram and use a variable to write an equation then solve)
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A................B........C..............D
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AB = BC + 15
CD = BC + 8
AD = 2*AB - 5
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AD = AB + BC + CD
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Substitute for AB and CD to get:
AD = BC+15+BC+BC+8
AD = 3BC+23
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Also: AD = 2*AB-5
So: AD = 2(BC+15)-5
And: AD = 2*BC+25
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Equation:
3BC+23 = 2BC+25
BC = 2
AB = BC+15 = 2+15 = 17
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CD = 2BC+23
So CD = 2*2+23 = 27
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Finally:
AD = AB + BC + CD
AD = 15 + 2 + 27 = 44
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Cheers,
Stan H.
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