What would be your answer to this math problem?

[Mom0012]

n/m - p = r; for n

 

I think our answer key is wrong, but it could be me.

 

Thx!

[FLDebbie]

n=(r+p)*m

[regentrude]

n= (r+p)*m

[Dana]

Yup...although key could have distributed and cleared parentheses. Depends on what you'll be doing which form is more useful. So also n= rm + pm

[Mom0012]

Yup...although key could have distributed and cleared parentheses. Depends on what you'll be doing which form is more useful. So also n= rm + pm

Ok, so this was the answer in the key.

 

What did we do wrong?

 

n/m - p =r; for n

 

m * n/m - p = r * m

 

n - p = rm

 

n - p + p = rm + p

 

n = rm + p

 

???

 

This seemed like such a simple problem, so what am I missing? There isn't any solutions guide and I would like to go over this with my son. I do see how you get the other answer if you start by adding p to each side, but how do I know to do that before multiplying each side by m? The teacher has emphasized getting rid of fractions first, so that's where we started.

 

Thx!

[Trez]

When you initially multiplied through by m, you have to multiply the whole left side by m. So you end up with:

 

m * (n/m - p) = r * m

 

Multiplying each term on the left by m you get:

 

n - pm = rm

 

n = rm + pm

[Mom0012]

When you initially multiplied through by m, you have to multiply the whole left side by m. So you end up with:

 

m * (n/m - p) = r * m

 

Multiplying each term on the left by m you get:

 

n - pm = rm

 

n = rm + pm

Thanks! I should have been able to see that, but we've just started to dip our toes into these types of problems and I remember *nothing* from high school math.

 

Interestingly, I was just reading a book last night that stated students who take math through calculus or higher remember algebra well even 55 years later. Apparently, there are exceptions, LOL.

[boscopup]

This seemed like such a simple problem, so what am I missing? There isn't any solutions guide and I would like to go over this with my son. I do see how you get the other answer if you start by adding p to each side, but how do I know to do that before multiplying each side by m? The teacher has emphasized getting rid of fractions first, so that's where we started.

 

It might help to see this with actual numbers:

 

 

8/2 - 3 = 1

 

Now, without reducing the fraction, try doing what you did first:

 

(8/2)*2 - 3 = 1*2

8 - 3 = 2

5 = 2

 

Clearly wrong.

 

Ok, so let's do it the right way... If you're multiplying the entire equation by a number, the equation has to all still be the same.

 

(8/2 - 3)*2 = 1*2

8 - 6 = 2    (distribute over multiplication)

2 = 2

 

Yay, we got it right!

 

See how the second time, we multiplied every component of the equation by 2? It kept the equation equal. 8/2-3=1 is the same as 8-6=2 is the same as 16-12=4.

 

Now most people moved the p over to the other side first because they knew they were going to be isolating n. If I were working with real numbers, I might want to get rid of the fraction first. It really depends on what I'm doing. So knowing that both methods are valid is important. I would have moved p over to the other side, then multiplied by m. Either way works though.

[Dana]

It might help to see this with actual numbers:

 

And this is a GREAT technique as you're learning how to solve literal equations or formulas for a given variable.

 

I'll show a problem with only one variable (like solve 3x - 5 = 6) on one side of the board and a formula (like Solve ax + b = c for x) on the other side of the board, and we'll do one step at a time with each problem... see how they're the same problem, but one has numbers and one has constants (or variables we're treating as constants).

 

The goal, of course, is to not need to put the numbers in, but as you're learning, it can be a useful help.

[Mom0012]

It might help to see this with actual numbers:

 

 

8/2 - 3 = 1

 

Now, without reducing the fraction, try doing what you did first:

 

(8/2)*2 - 3 = 1*2

8 - 3 = 2

5 = 2

 

Clearly wrong.

 

Ok, so let's do it the right way... If you're multiplying the entire equation by a number, the equation has to all still be the same.

 

(8/2 - 3)*2 = 1*2

8 - 6 = 2 (distribute over multiplication)

2 = 2

 

Yay, we got it right!

 

See how the second time, we multiplied every component of the equation by 2? It kept the equation equal. 8/2-3=1 is the same as 8-6=2 is the same as 16-12=4.

 

Now most people moved the p over to the other side first because they knew they were going to be isolating n. If I were working with real numbers, I might want to get rid of the fraction first. It really depends on what I'm doing. So knowing that both methods are valid is important. I would have moved p over to the other side, then multiplied by m. Either way works though.

This is very helpful. My ds is using TabletClass and the teacher has said to think of all the variables except the one you are trying to solve for as numbers, and I honestly wasn't sure why he was saying that. (I didn't hear that particular lecture, though. My son just told me that and I wondered why.) Your explanation makes it all much clearer to me.

[Farrar]

I also find it useful to substitute small numbers when you have a difficult problem with big, intimidating numbers (or just long, small ones).  It helps me get procedural clarity, if that makes sense.  As in, if doing it this way works for 2 and 10, then it will work for 1.4567 and 345 or whatever.

 

If you have an Ipad, I must say, Dragonbox has been helping me review this sort of thing with the distributive property and remember it to teach the kids. 

[Mom0012]

I also find it useful to substitute small numbers when you have a difficult problem with big, intimidating numbers (or just long, small ones). It helps me get procedural clarity, if that makes sense. As in, if doing it this way works for 2 and 10, then it will work for 1.4567 and 345 or whatever.

 

If you have an Ipad, I must say, Dragonbox has been helping me review this sort of thing with the distributive property and remember it to teach the kids.

Great idea about Dragonbox. I've already got it on my iPad for my kids, but I didn't think about using it for myself as a refresher. I also like what you are saying about substituting small numbers for larger ones. I think that will really resonate with my son.