a woodworking project calls for three pieces. the longest piece is twicw the length of the middle-sized piece, and the shortest is 10 inches shorther than the middle-sized piece. the board os 70 inches. how much must each peice be?
Let the middle-sized piece be x.
Then: 2x = longest piece and x-10 = shortest piece.
Therefore:
x + 2x + x - 10 = 70
Can you take it from here?
(Hint: solve for x, then be sure to list the length of the 3 pieces!)
SOLUTION: If the middle of three consecutive even integers is added to 100, the result is 42 more than the sum of the largest and twice the smallest. Find the smallest integer
20
To find the length of each piece, let's create variables to represent the lengths.
Let:
- L = Length of the longest piece
- M = Length of the middle-sized piece
- S = Length of the shortest piece
From the information given, we have three conditions:
1. The longest piece is twice the length of the middle-sized piece:
L = 2M
2. The shortest piece is 10 inches shorter than the middle-sized piece:
S = M - 10
3. The sum of the lengths of all three pieces adds up to 70 inches:
L + M + S = 70
Now, let's solve this system of equations to find the lengths of each piece.
Substituting equation 1 and 2 into equation 3:
(2M) + M + (M - 10) = 70
Combine like terms:
4M - 10 = 70
Add 10 to both sides of the equation:
4M = 80
Divide both sides of the equation by 4:
M = 20
Now that we have the value of M, we can substitute it back into equation 1 and 2 to find L and S.
Using equation 1:
L = 2M
L = 2(20)
L = 40
Using equation 2:
S = M - 10
S = 20 - 10
S = 10
Therefore, the lengths of the three pieces are:
Longest piece (L) = 40 inches
Middle-sized piece (M) = 20 inches
Shortest piece (S) = 10 inches