Let's set up the equations:
C=A+B+10 ....(1)
B=2A .....(2)
A+B+C=490 ....(3)
Substitute (2) in (1) to get:
C=A+2A+10=3A+10 .....(1A)
Substitute (1A) & (2) in (3) to get
A+2A+3A+10 = 490 ......(3A)
Solve for A and back-substitute in (2) and (1) to get B and C.
Hey I need some help with this problem. Any ideas on how to solve it? Thanks lots. : A hardware supplier manufactures three kinds of clamps, types A, B and C. Production restrictions require them to make 10 units more type C clamps than the total of the other types and twice as many type B clamps as type A. The shop must produce a total of 490 units of clamps per day. How many units of each type can be made per day?
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