How to 'change the subject' in algebra

MaddSci3ntisT · 2022-02-13 22:13:13

So this wasn't quite covered in detail. I'm trying to change the subject, and I'm completely lost. It's question 6 in equations and formulas. This is the question, https://imgur.com/a/fcme1Rr

& this is what I got
https://imgur.com/a/F38edAS

Where am I going wrong, or what else do I need to do?

Bob · 2022-02-13 23:04:19

hi MaddSci3ntisT

I've switched it round so you can ask yourself "What happens to b, and in what order?"

First b is doubled; then 2a is added.

'Undo' this by reversing these two steps going from right to left.

Then 'undo' the doubling.

Finally, write b =

A good way to get a thorough understanding and to check your answer is to substitute some numbers. The rules of algebra are the same as the rules of arithmetic.

eg. Let a = 3 and b = 7, then P = 2 x 3 + 2 x 7 = 6 + 14 = 20. Here you can see that b is doubled before the addition.

Is b = (P-2a)/2 correct?

Try those numbers.

b = (20 - 2x3)/2 = (20 - 6)/2 = 14/2 = 7. That's what b should be, so I'm probably correct.

I'm saying 'probably' because it is possible ( but unlikely ) that I've made two mistakes that 'cancel out' and still show b = 7.

That's why I chose 3 and 7; no common factors; and not 1 as times by 1 doesn't change anything.

Bob

ps. When I taught this at secondary level school (11-18) I devised a set of steps for changing the subject in virtually any formula. I'll post them if you would like.

MaddSci3ntisT · 2022-02-14 19:49:55

Bob wrote:

hi MaddSci3ntisT
I've switched it round so you can ask yourself "What happens to b, and in what order?"
First b is doubled; then 2a is added.
'Undo' this by reversing these two steps going from right to left.
Then 'undo' the doubling.
Finally, write b =
A good way to get a thorough understanding and to check your answer is to substitute some numbers. The rules of algebra are the same as the rules of arithmetic.
eg. Let a = 3 and b = 7, then P = 2 x 3 + 2 x 7 = 6 + 14 = 20. Here you can see that b is doubled before the addition.
Is b = (P-2a)/2 correct?
Try those numbers.
b = (20 - 2x3)/2 = (20 - 6)/2 = 14/2 = 7. That's what b should be, so I'm probably correct.
I'm saying 'probably' because it is possible ( but unlikely ) that I've made two mistakes that 'cancel out' and still show b = 7.
That's why I chose 3 and 7; no common factors; and not 1 as times by 1 doesn't change anything.
Bob
ps. When I taught this at secondary level school (11-18) I devised a set of steps for changing the subject in virtually any formula. I'll post them if you would like.

You are phenomenal, thank you for this response.

Yes, if you could post the steps for changing any formula, it would be greatly appreciated, as today is one of those 'colder' mornings where my vehicle is taking a while to start up C(=

In the meantime, I'm going to do as you advised and fool around on my whiteboard with the numbers example you gave me.

Bob · 2022-02-14 21:46:28

Ok. Here we go. For all 'linear' equations these four steps seem to always work. I'll come back later to what 'linear' means and what you do when an equation isn't linear.

Mostly you won't need all four steps so rather than throwing you in at the deep end I'll introduce them as needed.

The problem you started with needs steps 2 and 4.

Step 2: Get all 'b' containing terms on the left and everything else on the right.

So this step isolates the 'b' term on its own.

Step 4: Divide by the other factor for b.

In our case we have 2b and we want just b, so divide by 2

Here's another example. In Newtonian mechanics there's a formula:

Here v is the final velocity, u the initial velocity, a the constant acceleration, and t is time.

Let's make t the subject.

Step 2:

Step 4:

Here's some for you to try:

Q1. Using that formula make a the subject.

Q2. y = mx + c. Make c the subject.

Q3 Same formula, make x the subject.

Let's let that sink in and then I'll move on the step 3.

Bob

MaddSci3ntisT · 2022-02-15 01:25:47

Bob wrote:

Ok. Here we go. For all 'linear' equations these four steps seem to always work. I'll come back later to what 'linear' means and what you do when an equation isn't linear.
Mostly you won't need all four steps so rather than throwing you in at the deep end I'll introduce them as needed.
The problem you started with needs steps 2 and 4.
Step 2: Get all 'b' containing terms on the left and everything else on the right.
So this step isolates the 'b' term on its own.
Step 4: Divide by the other factor for b.
In our case we have 2b and we want just b, so divide by 2
Here's another example. In Newtonian mechanics there's a formula:
Here v is the final velocity, u the initial velocity, a the constant acceleration, and t is time.
Let's make t the subject.
Step 2:
Step 4:
Here's some for you to try:
Q1. Using that formula make a the subject.
Q2. y = mx + c. Make c the subject.
Q3 Same formula, make x the subject.
Let's let that sink in and then I'll move on the step 3.
Bob

I feel completely lost, I'm sorry. It took me a long while to try and grasp this, which, I think is actually quite simple. At least in comparison to THIS (the next question) https://imgur.com/a/REzt0Gq

I think this level of math might be a little beyond my reach right now. I did graduate from the grade 10-1 math university prep program back in 2015, here in Canada which is the equivalent of grade 10 math (so they say). I barely graduated. But this math is so mind boggling. I'll list all my questions to the next problem, I am NOT going to even think about clicking on the next question.

1. It says to multiply both sides of the formula by 9 to get rid of the fraction. The only math rule I have to go by for guidance is BEDMAS, and nowhere in that acronym does it specify how to do this type of math, either the fraction itself OR which component of the fraction, the numerator or the denominator. I'm assuming it's the denominator that always gets divided? What are the rules surrounding that operation? I would have thought I needed to start INSIDE the brackets, it's actually what I tried doing originally before just guessing the answer incorrectly.

2. Expand the right side, that one I can do. 5 multiplied by -32 is 160, 9C=5F-160

3. Add 160 to both sides. 9C+160=5F (I always get tripped up when I DON'T see what's being added to the other side, I tend to forget it cancels itself out)

4. Swap sides 5F=9C+160

5. Divide all terms by 5. How do I know I have to divide? And further, how do I know which number I have to divide by? I'm assuming because I had to multiply by 9 in step 1, that 5 has unfinished business with this equation and that's why we're here at this step right now?

I'm really sorry to spring this on you like this, I feel like equations and formulas are a little beyond me as of right now. Do you feel the same way? Is there another chapter I can go and learn before trying to do these sequences? Or is it just taking me an unusually long time to grasp this subject? I feel like there's lots of rules I'm not fully familiar with to do it properly.

Bob · 2022-02-15 03:11:46

I had that formula in mind for later, when you've got better with my simpler ones. In our car analogy that's for when you are in 4th gear and we're still just changing into 2nd.

An equation means the left hand side and the right hand side are equal. If you want them to remain equal you have to do the same to both sides. Here's a great animation that will show that in action:

https://www.mathsisfun.com/algebra/add- … lance.html

Each time you press 'new equation' you get another one to try. These only involve adding or subtracting. Multiplication is dealt with here: https://www.mathsisfun.com/algebra/intr … tiply.html ... sadly no animation this time.

I suggest you try both those pages and then my questions in the previous post. Once you can get those right you'll be ready to go up a gear.

My plan was steps 2 and 4 first because they are easier to follow. Then step 3 then step 1. That formula is for converting temperatures from fahrenheit to celcius. It requires step 1 which is last on my programme. (It has to be step 1 though, as you have to do it first when a problem requires it.

Bob

MaddSci3ntisT · 2022-02-15 22:47:31

Bob wrote:

I had that formula in mind for later, when you've got better with my simpler ones. In our car analogy that's for when you are in 4th gear and we're still just changing into 2nd.
An equation means the left hand side and the right hand side are equal. If you want them to remain equal you have to do the same to both sides. Here's a great animation that will show that in action:
https://www.mathsisfun.com/algebra/add- … lance.html
Each time you press 'new equation' you get another one to try. These only involve adding or subtracting. Multiplication is dealt with here: https://www.mathsisfun.com/algebra/intr … tiply.html ... sadly no animation this time.
I suggest you try both those pages and then my questions in the previous post. Once you can get those right you'll be ready to go up a gear.
My plan was steps 2 and 4 first because they are easier to follow. Then step 3 then step 1. That formula is for converting temperatures from fahrenheit to celcius. It requires step 1 which is last on my programme. (It has to be step 1 though, as you have to do it first when a problem requires it.
Bob

Hey my apologies, I'm trying to drive 50 in a 30 zone right now lol (good analogy btw).

So hopefully I got the gist of what's being asked. I'll upload the way I did the math via imgur and I'll also give my answers as well.

Q1 t=u-v/a
https://imgur.com/a/dLd5bZh

Q2 c=y-mx
https://imgur.com/a/q5N1Nbv

Q3 x=y-c/m
https://imgur.com/a/oxv8yGw

Bob · 2022-02-15 23:29:28

Ok. A bit of gear crunching but we're getting there.

Q2. Completely correct.

Q3. Correct on imgur. When you type it as text you need to surround y-c with a bracket to make it clear that the subtract takes precedence over the divide. The way you've written it is correct on imgur. As text it needs to be x = (y-c)/m

Q1 It was supposed to be 'a' the subject, not t. But also the same issue here. We want to force the subtraction first otherwise the PEDMAS rules (or do you say BODMAS?) would give the division precedence and your answer would be

rather than

If you try some numbers in the original to fix 'v' and then put v,u and t into each of the above you'll soon see which gives the correct a.

OK. Let's move onto step 3. It wasn't needed in any so far but it is in the next example.

eg. y + x = z + bx Make x the subject.

In this one there are two terms with x in them. If you just tried to get one of the x on its own and equal to something with everything else, then x would depend on an expression with x in. That's no use if we want a formula for calculating x.

x = z + bx - y is a correct formula but you cannot work out x from it as bx is part of the right hand side.

So we need a way to obtain a single x in the equation.

Step 2 says get all the x containing terms on the left. Up until now there was only one but now there's two.

Subtract y from both sides and subtract bx from both sides

Step 3: Factorise the x out of the left hand side expression.

Now we can apply step 4, divide by the other factor

And if I write it as text x = (z-y)/(1-b) brackets for the usual reason.

Here's some to try:

Q4. 2a + 2b = xa + yb Make b the subject.

Q5. t + r = 3t Make t the subject.

Q6. u = at - 2t Make t the subject.

Q7. 2h + hg = 7g - 9 Make g the subject.

In each case choose some numbers to make a correct equation at the start and then test out your answer to see if the numbers work out correctly.

Bob

MaddSci3ntisT · 2022-02-17 00:55:28

Bob wrote:

Ok. A bit of gear crunching but we're getting there.
Q2. Completely correct.
Q3. Correct on imgur. When you type it as text you need to surround y-c with a bracket to make it clear that the subtract takes precedence over the divide. The way you've written it is correct on imgur. As text it needs to be x = (y-c)/m
Q1 It was supposed to be 'a' the subject, not t. But also the same issue here. We want to force the subtraction first otherwise the PEDMAS rules (or do you say BODMAS?) would give the division precedence and your answer would be
rather than
If you try some numbers in the original to fix 'v' and then put v,u and t into each of the above you'll soon see which gives the correct a.
OK. Let's move onto step 3. It wasn't needed in any so far but it is in the next example.
eg. y + x = z + bx Make x the subject.
In this one there are two terms with x in them. If you just tried to get one of the x on its own and equal to something with everything else, then x would depend on an expression with x in. That's no use if we want a formula for calculating x.
x = z + bx - y is a correct formula but you cannot work out x from it as bx is part of the right hand side.
So we need a way to obtain a single x in the equation.
Step 2 says get all the x containing terms on the left. Up until now there was only one but now there's two.
Subtract y from both sides and subtract bx from both sides
Step 3: Factorise the x out of the left hand side expression.
Now we can apply step 4, divide by the other factor
And if I write it as text x = (z-y)/(1-b) brackets for the usual reason.
Here's some to try:
Q4. 2a + 2b = xa + yb Make b the subject.
Q5. t + r = 3t Make t the subject.
Q6. u = at - 2t Make t the subject.
Q7. 2h + hg = 7g - 9 Make g the subject.
In each case choose some numbers to make a correct equation at the start and then test out your answer to see if the numbers work out correctly.
Bob

Hey I'm kind of lost.

https://imgur.com/a/nPiaUcn

On that photo, when you said to subtract the terms you mentioned, I wrote it out like that. Is that correct? Is that how I was left with x - bx = z - y? And further, how do I know which terms are the correct terms to subtract? You said to subtract 'y' and 'bx' from both sides, how did you know which terms needed to be subtracted?

And as for factorise, I am not familiar with what that is or how to do it. How did we go from x - bx to x(1-b)?

Can you explain or link me the lesson in MIF that explains how to do that please?

I am very sorry to ask these sorts of things, it's very humiliating..

Bob · 2022-02-17 02:26:19

Step 2 separates the x terms from the non x terms. By subtracting y from both sides the 'y term' leaves the LHS and moves to the RHS. By subtracting bx from both sides the 'bx term' leaves the RHS and moves to the LHS.

So I've got the x terms on the left and everything else on the right. Step 2 achieved.

In our earlier examples we then had <factor> times x = something and so we could do 'divide by the <factor> to get x on its own. We cannot do that when there are two x terms. But x is a common factor of each term so it can be factorised outside a bracket.

Let's see that happening with numbers.

y + x = z + bx I'll choose y = 3, x = 5, b = 7 and then think about what I want z to be to make LHS equals RHS.

3 + 5 = z + 7 times 5 = z + 35 so I want z so that 8 = z + 35 so z = -27

So the equation is

3 + 5 = -27 + 7 times 5 or 8 = -27 + 35 = 8

Now what happens when I carry out step 2.

x - bx = z - y

5 - 7 times 5 = -27 - 3 -30 = -30 tick.

Now to factorise the x. When the x is factorised from the x term you cannot be left with nothing, so you have to think of it as

x = x times 1

Then when the x is factorised out , you're left with the 1

x - bx = x (1 - b) check the numbers 5 times 1 - 5 times 7 = 5 times (1 - 7) These are both worth -30.

Here's a link to factorising https://www.mathsisfun.com/algebra/factoring.html

These get harder than we need quite quickly (after the first example ) so I've looked elsewhere for some better practice questions. These look more like it ... first page only at this stage.

https://www.bbc.co.uk/bitesize/guides/z … revision/1

Bob

MaddSci3ntisT · 2022-03-02 14:21:46

Hey, sorry for my delayed absence. I was dealing with some other matters. I'm back now, ready to resume & I bought a fancy new whiteboard ^__^ maybe now I'll actually feel like a 'Mad Scientist' lol

So I think I did the first question correctly, I'm getting a little tripped up on factorizing, however. Take a look here https://imgur.com/a/4ZnmJ0B

But I do think I did it correctly.

Bob · 2022-03-02 20:36:21

Use the number substitution 'trick' to see if this works out. It doesn't because of one error near the start.

So you have those 'a' terms the wrong way round.

Bob

MaddSci3ntisT · 2022-03-02 23:53:57

Bob wrote:

Use the number substitution 'trick' to see if this works out. It doesn't because of one error near the start.
So you have those 'a' terms the wrong way round.
Bob

Can I use any number for my substitution trick? I know you mentioned using '3' and '7' because there is no common factors, but will any number suffice for this?

And when you said my 'a' terms are the wrong way around, would it have been right if I had it as 2a - xa as opposed to xa - 2a?

Bob · 2022-03-03 06:08:45

I want to choose values for a, b x and y that make this equation true. It has taken a while to find whole numbers for this and I had to have one negative.

a = -7, b = 1, x = 5, y = 23.

Let's check I've got those right.

2a + 2b = xa + yb becomes LHS = 2 times -7 + 2 times 1 = -14 + 2 = -12 RHS = 5 times -7 + 23 times 1 = -35 + 23 = -12

OK, That seems to have worked.

Now I'll try those numbers in your answer

b = (2a - xa) / (2 - y) = (2 times -7 - 5 times -7) / (2 - 23) = (-14 + 35) / -21 = 21 / - 21 = -1

It should have been +1 so something has gone wrong. but it's only wrong by a minus sign so switching xa and 2a around makes it right. It should be

b = (xa - 2a) / ( 2 - y)

Bob

Math Is Fun Forum

#1 2022-02-13 22:13:13

How to 'change the subject' in algebra

#2 2022-02-13 23:04:19

Re: How to 'change the subject' in algebra

#3 2022-02-14 19:49:55

Re: How to 'change the subject' in algebra

#4 2022-02-14 21:46:28

Re: How to 'change the subject' in algebra

#5 2022-02-15 01:25:47

Re: How to 'change the subject' in algebra

#6 2022-02-15 03:11:46

Re: How to 'change the subject' in algebra

#7 2022-02-15 22:47:31

Re: How to 'change the subject' in algebra

#8 2022-02-15 23:29:28

Re: How to 'change the subject' in algebra

#9 2022-02-17 00:55:28

Re: How to 'change the subject' in algebra

#10 2022-02-17 02:26:19

Re: How to 'change the subject' in algebra

#11 2022-03-02 14:21:46

Re: How to 'change the subject' in algebra

#12 2022-03-02 20:36:21

Re: How to 'change the subject' in algebra

#13 2022-03-02 23:53:57

Re: How to 'change the subject' in algebra

#14 2022-03-03 06:08:45

Re: How to 'change the subject' in algebra

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