Simple algebra problems can be expressed in terms of lengths of sticks. For example, 2x + 3 = 7 can be expressed this way: Suppose there are two sticks of equal length, and one stick of length 3. When they are placed end to end, they are the exact same length as a stick that has length 7.
I draw pictures. Unfortunately, that is not easy to do in computerland, so you will have to imagine. When there is just one unknown stick, I represent it with a question mark; with two or more sticks of the same length, I use letters. One example:

The simplest problem is ? = 3. All you have to do is realize that you have to use the second stick to know the length of the first stick. This can be used to create addition problems, as in ? = 4 + 7. Actually, the addition problems can be phrased in a way that is conceptually easier: "If you had a stick that was length 4 and a stick that was length 7, and you put them end to end, how long would they be?" If a student could not get ? = 3 with two sticks, you could start out by asking how long the combined length is. Then the next problem could be, how long is a stick next to it if the new stick is the same length as 4 and 7 put together.

These can be given to young children. You have to pick numbers to suit a child's skill. My problems for 3rd graders:

Set 1: ? + 3 = 9 + 3 ? + m = 17 + m ? + 108 = 62 + 108. I explain for the second problem that the length of the sticks marked m is unknown, but they are the same length.

Set 2: 7 + 1 = ? 7 + ? = 8 ? + 1 = 8

Set 3: 6 + 5 = ? 2 + 5 + 8 = ? Simple addition.

Set 4: ? + 5 = 8 3 + ? = 10 problems in the form of x + a = b, using simple numbers

Set 5: ? + 15 = 24 3 + ? = 10 problems in the form of x + a = b, using larger numbers. Now the answer cannot be seen.

Set 6: ? + 2 = 95 A visual joke -- I drew the 2 big and the unknown stick as small. No one noticed.

Set 7: 3 + 4 + ? 9 6 + ? + 5 = 20 ? + 92 + 63 = 205 problems in the form of x + a + b = c. Requires simplification.

Set 8: ? + 17 = 23 + 29 ? + 21 + 25 = 35 + 42 The first problem requires simplification of the other side of the equation. The second problem requires simplication of both sides.

Set 9. m + m + m = 9 (I will call this 3m = 9) 2m = 40 3m = 39. I labelled the unknown stick length as m. Again, the convention is that two sticks marked with the same letter have the same length. But now the problem is to discover m. I had previously worked with variables.

Set 10. 8m = 32 9m = 621 Larger numbers still using the format ax = b

Set 11. 2m + 1 = 5 2m + 5 = 17 m + 39 + m = 81 These are problems of the form ax + b = c. This set goes too fast -- the first problem is small numbers where the student can see the answer, the second problem gets larger, and the third problem is still bigger numbers and has the concept of combining m's that aren't right next to each other.

Set 12: 8m + 12 = 68 6m + 16 = 46 using bigger numbers. This makes addition not as feasible so they have to rely on division.

Set 13 2m + 3 = m + 5 2m + 13 = m + 18 3m + 6 = 2m + 8 More complexity. Now there are m's on the other side.

Set 14: 3m + 12 = 2m + 15 m + 9 + m = 11 + m 2m + 9 = 11 + 3 More of this style. In the third problem, the m's are not matched up in position.

Set 15: meant to be funny. m + 0 = 17 (I drew a stick with size and labelled it as zero.) 12 + m = m + 12 (unsolvable) 3m + 17 = 17 (Again, the sticks marked m had size, so the two sticks marked 17 were different lengths) 5 * m = 5 * 62 (This should be solved without multiplication)

Set 16: m + 3 + m = 2 + m + 2 m + m + 3 = m + 2 + 2 m + 7 + m = 8 + m + 8 no new concept? Again, m's are misaligned.

Set 17: 6m + 14 + 1 = 8 + m + 6 + m + 3 + m + 10 Requires serious simplification. This in a sense is the ultimate complexity of a 1 variable problem.

Set 18: Can add another unknown stick, as long as it appears an equal number of times on each side of the equation.

Variations: These problems are all adaptable to weights on a scale. Some of the simpler ones can be the number of boys and girls in two classrooms where each classroom has the same number of students. Time is also additive. So in theory any of these could be a time story problem.

Variations: Have the student write out the formula that is being tested. Have the student convert a formula to a stick problem.

Obviously: Obviously, an eventual step is to teach children to solve equations. I suspect there is no reason to be in any hurry to do that. Too often, solving equations is the mindless manipulation of symbols. Bob sez to become very comfortable with these. Then solving equations would be easy to learn.

### Two Equations with two unknowns

I suggest being very facile with simplification of complex one-equation problems before trying these. I think students also probably need to know the silver and gold problems.

Set 1: x + y = 18, x = y x + y = 19, x = y + 1