Question 1124862
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(1) An informal solution , using logical reasoning and a bit of mental arithmetic....<br>
Make those 10 "extra" type C clamps first.
That leaves 300 more clamps to make; and of those the number of type C clamps is the same as the total of the types A and B clamps.
That means 150 more type C clamps (for a total of 160), and a total of 150 clamps of types A and B.
Then twice as many type B as type A means 100 type B and 50 type A.<br>
Answer: 50 type A, 100 type B, 160 type C.<br>
Using formal algebra....<br>
(1) A+B+C = 310
(2) C = A+B+10
(3) B = 2A<br>
There are many algebraic paths from these three equations to a solution.  Here is one -- which, it turns out, doesn't look much at all like the informal solution above.<br>
Using (3), substitute 2A for B in (1) and (2):
(4) 3A+C = 310
(5) C = 3A+10<br>
Using (4), substitute  3A+10 for C in (5):
(6) 3A+3A+10 = 310<br>
Solve for A:
6A+10 = 310
6A = 300
A = 50<br>
Substitute A=50 in (5):
150+C = 310
C = 160<br>
Substitute A=50 and C=160 in (1):
50+B+160 = 310
B = 100<br>
You can see that, in this problem, the informal solution is much easier than the one using formal algebra....