translate the given information into a system of equations and solve it.

Question

a company manufactures three types of hammers good better,and best.the cost of manufacturing each type of hammer is P4, P6 and P7,respectively,and the hammers sell for P6, P9 and P12.Each day cost of manufacturing 100 hammers is P520 and the daily revenue from their sale is P810.How many of each type are manufactured?

Answer 1

x is the # of good hammers

y is the # of better hammers

z is the # of Best Hammers

 

X+y+z  = 100  <--- there are 100 altogether

 

4x+6y+7z = 520  <---- the total cost to produce them is 520 at the given cost for each

 

6x+9y+12z = 810  <--- the total revenue is 810

 

 

Equation 2 - 4 * equation 1: 2y + 3z = 120

equation 3 - 6* equation 1 : 3y + 6z = 210

 

 

Solves by eliminations:  -3 times first equation and 2 times second equation

   -6y + -9z = -360

   6y  + 12z = 420

 

   3z = 60

   z = 20   ---->   6y + 12(20) = 420

                         6y  + 240 = 420

                          6y = 180

                           y = 30

 

So then x = 50