Solving a word problem using linear equations

Question

A fruit company delivers its fruit in two types of boxes: large and small. A delivery of 2 large boxes and 12 small boxes has a total weight of 199 kilograms. A delivery of 5 large boxes and 3 small boxes has a total weight of 133 kilograms. How much does each type of box weigh?

Answer 1

Let b = the weight of one small box, and

let B = the weight of one large box.

Then we have two simultaneous equations to solve:

2B + 12b = 199

5B + 3b = 133

To solve, multiply the 2nd equation by 4 to get

20B + 12b = 532

Now subtract the third equation from the first to get

-18B = -333

Solve for B:

B = -333 / -18 = 18.5

We can use any of the first three equations to solve for b; let's use the simplest:

5B + 3b = 133

b = (133 - 5B) / 3 = (133 - 5*18.5) / 3 = 13.5 kgs

Check:

2B + 12b = 2(18.5) + 12(13.5) = 37 + 162 = 199

5B + 3b = 5(18.5) + 3(13.5) = 92.5 + 40.5 = 133

Answer 2

Delivery 1

2 large, 12 small, 199kg

2L + 12S = 199

Delivery 2

5 large, 3 small, 133kg

5L + 3S = 133

1. Solve for L
2. Make S removable from the equation. Since variable S can become equivalent to the S in the other equation, S can become cancelled.
3. Delivery 1 has 12S, Delivery 2 has 3S.
4. Multiple Delivery 2's equation by 4
5. 4 (5L + 3S) = 4 (133)

20L + 12S = 532

1. Remove S by subtracting one equation from another
2. 2L + 12S = 199

20L + 12S = 532 (Delivery 2 Equation multipled by 4)

-2L - 12S = -199 (Delivery 1 Equation)

18L = 333

L = 333/18

L = 18.5

1. Solve for S, using L=18.5
2. 2 (18.5) + 12S = 199

37 + 12S = 199

12S = 199 - 37

12S = 162

S = 162/12

S = 13.5

1. Plug in both answer to check your work. If plugging in your answers for L and S solves the equation correctly, you have proven your work.
2. 2(18.5) + 12(13.5) = 37 + 162 = 199
3. Correct
4. 5(18.5) + 3(13.5) = 92.5 + 40.5 = 133
5. Correct

Large boxes = 18.5kg

Small boxes = 13.5kg

Answer 3

You can begin this problem by rewriting the word problems in the "system of equation" form using variables L (for Large) and S (for small).

Thus,

2L + 12S = 199

5L + 3S = 133

_____________

Now you need to figure out how to cancel one of the variables so that you can solve for the other.

Let's cancel S.

If we multiply the entire second equation by -4,

-4(5L + 3S = 133)

This equation becomes,

-20L - 12S = -532

Now, to solve the system, we combine each term in the 2 system equations vertically.

So,

2L + 12S = 199

-20L - 12S = -532

_______________

This cancels out S (since 12S-12S = 0)

and we have: -18L = -333

Dividing by -18 on both sides yields L= 18.5

Now we can take the L we just found and plug it into either one of the original equations.

Let's plug it into 2L + 12S = 199

We get: 2(18.5) +12S = 199

This simplifies to: 37 + 12S = 199

Subtracting 37 from both sides: 12S = 162

Dividing by 12 on both sides: S = 13.5

Therefore, the small boxes (S) weigh 13.5 kg and the large boxes (L) weigh 18.5 kg