More Problem Solving For Little Kids

Basically, we are playing advanced "what's under the cup?" but the name of this game is "finding stuff that's same on both sides."

It's fun and even easier than "what's under the cup?" if you do it right. Start off easy and all positive and no fractional answers.

The answers to problems like these are visually obvious. Once you remove the stuff that's same on both sides, this is a slightly more advanced problem because it involves hero zero AND no fun get back to one. The boys can see the answers and what to do. It's FUN when you get it right. Some observant people have noted there are 5x worth of manipulatives on the one side instead of 4, don't worry we sort it out in the longer video.

These two had been couped-up in the house for a couple of days with runny noses and fever but were still able to play math a little bit. Along with several hellacious games of Chutes and Ladders...

Here is the short version where all they do is set up this problem, in the full video we solve this one and several more. The boys say they are easy-peezy. I know several high school students who would beg to differ...as well as a few home school moms.



4x + 1 = 2x + 9

Child's play. This problem is easy if you can see it. Many people I know tend to have their eyes glaze over as soon as they see the algebra. Once they see the problem solving page at Crewton Ramone's House Of Math they can't believe it's that easy. Really, most of the math is child's play and you can make games of it if you are willing to get creative. There are math teachers who get quite excited when they see this way of teaching. Parents who thought they couldn't do math also get excited but they lack the knowledge to make up problems easily on their own, this can be remedied with practice. Here are a couple of story problems that can be represented with these symbols:

4x + 1 = 2x + 9

Bob and Jay are out of town. Bob is trying to catch Jay. Bob is on his skate board going 4 miles an hour, Jay is walking along at 2 miles an hour. Bob is one mile out of town and Jay is 9 miles out of town. How long does it take Bob to catch Jay and how far out of town are they when he does?

Or you could make it a little more difficult: Jay is walking at 2 miles an hour and is 9 miles out of town. Bob is only 1 mile out of town but is walking twice as fast. How long does it take Bob to catch Jay and how far out of town are they when he does?

Same Problem different story. Bob and Jay are going to have a snow ball fight. It the middle of summer in Utah and there is still over 8 feet of snow on the ground in places that are usually green with grass growing on the ground. How long before Jay and Bob learn enough math to debunk global warming claims? No wait...let's try that again:

Bob and Jay are going to have a snow ball fight. Bob has 1 snow ball and Jay has 9, which isn't fair. So they each build more snowballs. Jay works for 2 minutes and Bob works for 4 minutes, they build the same amount of snowballs per minute and when they are done they each have the same amount of snow balls. How many snow balls do they have and how many did they make each minute?

Be sure to check out the previous blog post on problem solving for more. And if you want to see the entire video get yourself a password and go here.