SOLUTION: will someone help me please? When the circus came to town last year, they hired general labors at $70 per day and mechanics at $90 per day. They paid $1950 foe this temporary help

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: will someone help me please? When the circus came to town last year, they hired general labors at $70 per day and mechanics at $90 per day. They paid $1950 foe this temporary help      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 731664: will someone help me please?
When the circus came to town last year, they hired general labors at $70 per day and mechanics at $90 per day. They paid $1950 foe this temporary help for one day. This year they hired exactly the same number of people of each type, but they paid $80 for general labors and $100 for mechanics for one day. This year they paid $2200 for temporary help. How many general laborers did they hire? How many mechanics did they hire?
thanks !! ;)
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
x= number of general laborers
y= number of no of mechanics

70.00 x + 90.00 z = 1950.00 .............1
Total value
80.00 x + 100.00 z = 2200.00 .............2
Eliminate y
multiply (1)by -10.00
Multiply (2) by 9.00
-700.00 x -900.00 z = -19500.00
720.00 x + 900.00 z = 19800.00
Add the two equations
20.00 x = 300.00
/ 20.00
x = 15.00
plug value of x in (1)
70.00 x + 90.00 z = 1950.00
1050.00 + 90.00 z = 1950.00
90.00 z = 1950.00 -1050.00
90.00 z = 900.00
z = 10
x= 15 number of general laborers
y= 10 number of no of mechanics
m.ananth@hotmail.ca