SOLUTION: 1.5 hours and the material costs $8. Method II takes 2.0 hours, and the material costs $5 Next week, they plan to spend 158 hours in labor and $480 in material for refinis

Algebra ->  Systems-of-equations -> SOLUTION: 1.5 hours and the material costs $8. Method II takes 2.0 hours, and the material costs $5 Next week, they plan to spend 158 hours in labor and $480 in material for refinis      Log On


   

Question 1152850: 1.5 hours and the material costs
$8. Method II takes
2.0 hours, and the material costs
$5 Next week, they plan to spend
158 hours in labor and
$480
in material for refinishing tables. How many tables should they plan to refinish with each method? (Round to two decimal places if necessary.)
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

x tables with Method I.
y tables with Method II.
account for hours:
1.5x%2B2y=158
account for cost:
8x%2B5y=480
solve the system:
1.5x%2B2y=158.......1
8x%2B5y=480..........2
--------------------------------------
1.5x%2B2y=158.......1...solve for y
y=158%2F2-1.5x%2F2
y=79-0.75x.....1a ..substitute in eq.2
8x%2B5%2879-0.75x%29=480..........2
8x%2B395-3.75x=480
4.25x=480-395
4.25x=85
x=85%2F4.25
x=20
y=79-0.75x.....1a
y=79-0.75%2A20
y=79-15
y=64

so, they should plan
20tables with Method I.
64 tables with Method II
check:
20tables with Method I.:
20%2A1.5=30 hours and the material costs $20%2A8=160

64 tables with Method II:
64%2A2.0=128 hours, and the material costs $64%2A5+=320

=>30%2B128=158 hours and $160%2B320=480 cost