How do I solve this algebraically (without trial and error)

Question

A moving company weighs 22 boxes you have packed that contain either books or clothes. The total weight of these boxes is 445 pounds. If each box weighs 35 pounds and each box weighs 10 pounds, how many boxes of books did you pack?

Answer 1

Let x = the number of boxes of books. Since there are 22 boxes in total, the number of clothing boxes is 22 - x.

Total Weight = 35x + 10(22-x)

445 = 35x + 220 - 10x

Solve for x.

Answer 2

You can create a system of equations with the given information. Now you didn't specify, but I'm going to guess that each box of books is 35 lbs. and each box of clothes is 10 lbs. If this assumption is incorrect, just switch the symbols used below.

You know that the total number of boxes you have is 22, so if we use b for the number of boxes of books and c for the number of boxes of clothes, we can set up an equation to reflect this information:

b + c = 22

The second piece of information that we know is the weight of each box of each type. If we take the number of boxes of books and multiply that by the weight per box of books, we can get the total weight of just the boxes with books. You can use the units to see how this would look...

Before we get there, just to quickly clarify, 35 lbs PER box of books can be shown like this:

35 lbs ⁄ box of books

This is because the word "per" means that you are dividing the weight by a single box of books. This is identical to how 65 miles PER hour can also be shown like this: 65 miles ⁄ hour.

With that, we can now see that multiplying the number of boxes of books, b, by the weight of each box of books, and multiplying the number of boxes of clothes, c, by the weight of each box of clothes, we can get the total weight.

[(35 lbs ⁄ box of books) × b] + [(10 lbs ⁄ box of clothes) × c] = 445 lbs of total weight

You now have 2 equations with 2 variables, and so you can algebraically solve for each variable.

{

b + c = 22

35b + 10c = 445

}

You can solve the system whichever way you prefer, either through substitution or by adding / subtracting the equations to eliminate a variable. In this case, substitution is probably easier...

b = 22 - c

Solve first equation for b.

[35 × (22 - c)] + 10c = 445 lbs

Substitute what you got for b in the first step into the second equation.

[35×22 - 35c] + 10c = 445 lbs

Distribute the 35 to both terms inside the parentheses.

770 - 35c + 10c = 445

Multiply 35 times 22 and simplify the left side of the equation.

-35c + 10c = 445 - 770

Group like terms by subtracting 770 from both sides of the equation.

-25c = -325

Add like terms together.

c = +13

Divide by -25 and remember that a negative divided by a negative is positive.

So the number of boxes with clothes is 13 boxes. By plugging that value into the first equation, we can see that the number of boxes of books would be 9 boxes of books.

Let's check those values:

b + c = 22

9 + 13 = 22

This checks out.

35b + 10c = 445

35(9) + 10(13) = 445

315 + 130 = 445

This also checks out.

Therefore, we have our two value of the number of boxes of each type: We have 13 boxes of clothes and 9 boxes of books.