SOLUTION: I am thrown by this one. I have no total to work with and I should read it some other way. Please help.
This is how I started.
placing planks of equal length end to end, J foun
Algebra.Com
Question 52092This question is from textbook PH Alg-1
: I am thrown by this one. I have no total to work with and I should read it some other way. Please help.
This is how I started.
placing planks of equal length end to end, J found that 3 planks were one foot too short and 4 planks were two feet too long. Find the length of the planks
I did: X+3(X-1)+4(X-2)
where am I wrong?
This question is from textbook PH Alg-1
Found 2 solutions by checkley71, Earlsdon:
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
LETS TRY A DIFFERENT APPROACH AND DO IT GRAPHICLY
PLANK1+PLANK2+PLANK3 IS SHORT BY 1 FOOT OR -1
PLANK1+PLANK2+PLANK3+PLANK4 IS 2 FEET LONG OR +2 THE DISTANCE BETWEEN -1&+2=3
THEREFORE THE LENGTH OF THE PLANK MUST BE 1 FOOT + 2 FEET=3 FEET LONG
SORRY FOR THE MISINFORMATION. I MUST HAVE HAD A LAPSE BECAUSE I KNEW THE ANSWER HAD TO BE 3 FEET BUT DIDN'T PAY ATTENTION TO THE ANSWER I GOT.
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Try this:
Let x = the length required to be covered and L = the length of a plank.
1) 3L = x-1 Multiply this by 4.
2) 4L = x+2 Multiply this by 3.
1a) 12L = 4x-4
2a) 12L = 3x+6 Subtract equation 2a) from 1a)
3) 0 = x-10 Add 10 to both sides.
x = 10 Substitute this into equations 1) and 2) and solve for L
3L = 10 - 1 = 9: L = 3
4L = 10 + 2 = 12: L = 3
The planks are 3 feet long.
Check:
3L = 9 = 10-1
4L = 12= 10+2
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