Hi I am in desperate need of a "tutor" If anyone is decent at Algebra II, any little help is welcomed! Please respond to this an
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Answer:
The number of tablets 200
the number of laptops 40
and, the number of servers 20
Step-by-step explanation:
Let the number of tablets be 'x'
the number of laptops be 'y'
and, the number of servers be 'z'
According to the given question
Total budget = $280,000
Tablets cost = $500 each
laptops cost = $2000 each
servers cost = $5000 each
Thus,
500x + 2000y + 5000z = 280,000 ..............(1)
also,
x = 5y ................(2)
and,
y = 2z
or
z = ............(3)
substituting y from 2 and 3 in equation 1, we get
(500 × 5y) + (2000 × y) + (5000 ×
) = 280,000
or
2500y + 2000y + 2500y = 280,000
or
7000y = 280,000
or
y = 40
substituting y in equation 2, we get
x = 5 × 40 = 200
substituting y in equation 3 we get
z =
or
z = 20
hence,
the number of tablets 200
the number of laptops 40
and, the number of servers 20