A company manufactures three types of cabinets. It makes 110 cabinets each week. In the first week, the sum of the number of type-1 cabinets and twice the number of type-2 cabinets produced was 10 more than the number of type-3 cabinets produced. The next week, the number of type-1 cabinets produced was three times more than in the first week, no type-2 cabinets were produced, and the number of type-3 cabinets produced was the same as in the previous week.
In the first week, the number of type-1 cabinets produced was , the number of type-2 cabinets produced was , and the number of type-3 cabinets produced was .

Type cabinet 1 answer - 10 , 15 , 25 ,30
Type 2 - 20 , 25 , 30 , 45
Type 3 - 35 , 45 , 55, 65

WHEN YOU COMMENT PLEASE BE ONE HUNNA PERCENT!! AND JUST GIVE ME THE ANSWER

If we let a, b, c represent the numbers of type-1, type-2, and type-3 cabinets produced in the first week, respectively, we can write three equations in these unknowns:

a + b + c = 110 . . . . . total cabinets for the first week
a + 2b - c = 10 . . . . relationship of quantities in the first week
3a +0b +c = 110 . . . . total cabinets in the second week

It can be convenient to let a machine solver find the solution to this set of equations. Most graphing calculators can handle it, along with several web sites.

__

Solving by hand, we can subtract the second equation from twice the first. This gives ...

2(a +b +c) -(a +2b -c) - 2(110) -(10)

a +3c = 210 . . . . simplify

Subtracting this from 3 times the third equation gives ...

3(3a +c) -(a +3c) = 3(110) -(210)

8a = 120 . . . . . simplify

a = 15 . . . . . . . divide by 8

Using this in the third equation of the original set, we have ...

3·15 +c = 110

c = 65 . . . . . . subtract 45

Then, in the first equation, we get ...

15 + b + 65 = 110

b = 30 . . . . . . . subtract 80

The solution is (type-1, type-2, type-3) = (15, 30, 65) for the first week.