asks if there is a clean cup in the cupboard, and if no, then it says to clean a cup. It doesn’t say where to look for a clean cup if there isn’t one in the cupboard. It also doesn’t suggest where to find a cup to clean. So if ‘no’ I think the flow chart would stall without using creativity. Outcome = no cup of tea
no teabags and no milk
if you decide you don’t want to end the process when you get a no, then you’ll go to the store two separate times. You’ll check for tea bags, go to the store and get some, then check for milk, then go to the store and get some. That’s just inconvenient and inefficient. It wouldn’t stall the flowchart though and you would still finish it without creativity. No optimal though. Outcome = cup of tea
no milk and no clean cup
This has the same issue as the no clean cup. You have to use creativity to find a cup to clean, or to find an already clean cup elsewhere. Outcome = no cup of tea
no clean cup and no tea bags
Same issue about clean cup. Outcome = no cup of tea
power outage
doesn’t really handle it. The kettle switch may flick on when you set it to boil but you’ll be stuck at ‘has the kettle finished boiling?’ until the power comes back on. Outcome = no cup of tea.
no running water
wont get past first step. Outcome = no cup of tea
kettle is broken
you’ll be stuck at ‘waiting to boil’. Outcome = no cup of tea
there are no cups (e.g the last one broke)
kinda similar to the no clean cups problem. you’ll never find one unless you go to the store to get one. Outcome = no cup of tea.
Sure. Here is an improved flow chart. I’ve also made the question explicit about whether you’re willing to go to the shops right now in order to make a cup of tea.
Yeah okay. It’s obviously much simpler. Mine does looks pedantic and over-complex in comparison. Your doesn’t handle problems in the process that mine can and yours doesn’t have any decisions.
I guess when I thought about the process certain problems came up that I thought I shouldn’t just assume would be solved by background knowledge. Kinda like writing an algorithm. So am I including too much detail?
I’m finding this quite hard to put into words such that it applies in general. My flowchart feels too simple. But I didn’t want to get stuck on this for long.
there is only one term that contains the variable. I suppose this step could be rewritten as ‘reduce to one the number of terms that contain the variable if there are more than one’.
isolate the variable by moving terms to the other side of the equation by undoing their operations
Ok. subtract 1 from both sides then divide both sides by 9.
example problem: (9x + 1)/2 = 5x
combine terms that contain the variable
hmm. I don’t think there is a way to do this in one step. You’d have to first simplify the left expression, or multiply the right expression by 2/2 so they have the same denominator.
I feel a bit stuck and it’s causing me some frustration/distress.
Subtract (from both sides of the equation) any left-side constants. (Do the most basic simplification as you go. If there’s a 4 on the left, and you subtract 4, remove it instead of having 4-4).
Subtract any right-side terms with the variable.
Multiply by any denominators on the left.
Combine terms on the left (using addition and subtraction).
There’s now one variable term and its on the left. Divide by its coefficient.
Simplify the right side.
The result is x = n/m (where x is your variable and n and m are integers)
Subtract (from both sides of the equation) any left-side constants. (Do the most basic simplification as you go. If there’s a 4 on the left, and you subtract 4, remove it instead of having 4-4).
Subtract any right-side terms with the variable.
Multiply by any denominators on the left.
Combine terms on the left (using addition and subtraction).
There’s now one variable term and its on the left. Divide by its coefficient.
Simplify the right side.
The result is x = n/m (where x is your variable and n and m are integers)
Example problem: 3c + 6 = 18
Step 1: subtract 6:
3c = 12
Steps 2, 3 and 4: none:
3c = 12
Step 5: divide by 3:
c = 12/3
Step 6: simplify:
c = 4
OK.
Example problem: 8 + 2/z = 10
Step 1: subtract 8:
2/z = 2
Step 2: none:
2/z = 2
Step 3: multiply by z:
2 = 2z
Step 4: none:
2 = 2z
Step 5: There is no variable on the left side any longer since step 3.
Not OK? I think it’d work if it didn’t specify which side the variable was on in step 5 and which side to simplify in step 6?
That’s not a linear equation because of the term 2z^-1
Linear equations raise variables to the 0 and 1 powers only.
To see the issue visually with x^-1 terms, go to wolfram alpha and enter these:
graph y = 3x
graph y = 3/x
As an equation with non-linear features, your example is a special case where it could be converted to a linear equation. Other examples like that include x^(1/2)=3 or y^3=8.
Oh okay cool. I can see that they don’t make straight lines on the graph. I don’t fully understand why. In the case of x^-1 it seems to have something to do with dividing by zero.
I’ll do some more problems with linear equations with your steps.
Yeah I think so. Here’s my understanding of the steps you gave.
Subtract (from both sides of the equation) any left-side constants. (Do the most basic simplification as you go. If there’s a 4 on the left, and you subtract 4, remove it instead of having 4-4).
Step 1 means the left of the equation only has variable terms
Subtract any right-side terms with the variable.
Step 2 means we only have variable terms on the left of the equation
Multiply by any denominators on the left.
Step 3 gets rid of denominators on the left so you can then combine multiple instances of variables
Combine terms on the left (using addition and subtraction).
Step 4 leaves us with one variable term
There’s now one variable term and its on the left. Divide by its coefficient.
Step 5 tell us what one unit of the variable is equal to i.e what the solution is
Simplify the right side.
Step 6 gives us the solution in the simplest form
So I think basically all the steps are there. I would add that separating the terms step, so:
