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Answer :
Final answer:
Following the given relations and equations among Xavier, Yosef, and Zeke's push-ups, we calculate that Yosef does 20 push-ups.
Explanation:
The problem involves solving a series of equations to find out how many push-ups Yosef can do. We are given that Xavier, Yosef, and Zeke together do 56 push-ups in 3 minutes. Xavier does as many push-ups as Yosef and Zeke combined, and Zeke does two less than half as many push-ups as Yosef.
Let's denote the number of push-ups Yosef does as Y, Zeke does as Z, and Xavier does as X. From the information, we can set up the following equations:
- X = Y + Z (Xavier does as many as Yosef and Zeke combined)
- Z = (1/2)Y - 2 (Zeke does two less than half as many as Yosef does)
- X + Y + Z = 56 (Together, they do 56 push-ups)
Substituting equations 1 and 2 into equation 3 gives us:
Y + ((1/2)Y - 2) + (Y + (1/2)Y - 2) = 56
Simplifying, we find that 3Y - 4 = 56, leading to 3Y = 60, and therefore Y = 20.
So, Yosef does 20 push-ups.
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Rewritten by : Barada
Yosef did 20 push-ups
Xavier(x), Yosef(y), Zeke(z).
x+y+z=56
y+z=x
(y/2)-2=z
we plug in for z to solve for y
y+(y/2)-2=x
Combine like terms
1.5y-2=x
Plug in for x and z in terms of y in the original problem
1.5y-2+y+.5y-2=56
Add like terms
3y-4=56
Add 4 from both sides
3y=60
Divide both sides by 3
y=20
Xavier(x), Yosef(y), Zeke(z).
x+y+z=56
y+z=x
(y/2)-2=z
we plug in for z to solve for y
y+(y/2)-2=x
Combine like terms
1.5y-2=x
Plug in for x and z in terms of y in the original problem
1.5y-2+y+.5y-2=56
Add like terms
3y-4=56
Add 4 from both sides
3y=60
Divide both sides by 3
y=20