Doodles Do Algebra – Lesson 6


Update: This Lesson Is Part of Book 1 of Doodles Do Algebra: “Starting Out With Mental Algebra” available on Amazon.


Teacher’s Notes:

Today’s algebra lesson expands on the last 5 and adds a new complexity: figuring out quantities of two items when they are not the same but are related by some constant (like one being twice the other or one being three less than the other…)

This is where it becomes helpful to think in terms of x’s, but the problems can still be generally done in your (or your child’s head). For those of you who remember algebra class, this is the if you have two independent unknowns, you need two independent equations to solve for the unknowns” maxim at work. Last week all the problems were “one independent unknown with one independent equation” type, even if it didn’t always seem like it, since each problem collapsed down into one unknown even if we were talking about two people doing something. They were always equal distances away from each other, or an equal number of fruits were purchased, or the stick was broken in the middle. Today we look at how to solve the problems where things are not equally divided. It isn’t hard, as you might imagine, just different.



Available in the book, “Starting Out With Mental Algebra, Book 1 of Doodles Do Algebra”



  1. We have two numbers, that are not equal but one is twice the other, that when you add the larger to twice the smaller, the sum is 28. If you are reading this in a quiet moment, and are fully awake and caffeinated, you will notice that the larger number is 2 times the smaller and you add the larger (2 times the smaller) to the twice the smaller (2 times the smaller again!) to get a total of 28. That means you are adding 2 times the smaller to 2 times the smaller, so you get 2+2 (think in terms of 2 x’s plus 2 x’s makes 4 x’s, or apples or pennies just like last week), or 4 times the smaller equaling 28, or the smaller number being 7 and the larger, twice the smaller, or 14. So the smaller number is 7, larger number is 14.
  2. Now we do the same thing as in the first problem but instead of numbers we are talking about the price of oranges and lemons. Auntie 2 oranges and 5 lemons, and spent 27 cents. The second and most important fact you are given is that an orange costs twice as much as a lemon (last week, this problem would have been simplified to lemons and oranges costing the same, but this week we are taking off the training wheels!) So whatever you spent on a lemon, let’s call it x cents, you spent twice that, or 2x on an orange. Now you add up 5 lemons at x cents (or 5x) plus 2 oranges at 2x cents (or 4 x) equals 27 cents total. So 4x plus 5x equals 27, or 9x equals 27, or x equals 3 cents. Now you go back to the definition of lemons costing x cents and oranges costing twice that or 2x cents. So a lemon costs 3 cents and an orange costs 6 cents.
  3. Alright, now we do it one more time only this time the unknown is the number of fruit, not the price. Crockett (who is our big black standard poodle) bought apples and peaches. He bought twice as many apples as peaches so x is the number of peaches and 2x is the number of apples. He spend a whopping 24 cents on the whole pile of fruit, and apples cost 2 cents and peaches cost 4 cents each. So Crockett bought x peaches at 4 cents (or 4x) plus 2x apples at 2 cents (or 4x) and that added up to a total price of 24 cents. So 4x plus 4x equals 24, or 8x equals 24, and so x equals 3 fruits. Now we go back to our definition of twice as many apples purchased as peaches, so Crockett bought 3 peaches and 6 apples.

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