What basic algebra fact am I forgetting?

Factual Questions
1

I’m so rusty! I’m trying to understand the moves from equation #1 to equation #2 and I’m confused. In moving the (-16.0 m/s2) to the left side, I figure that I can subtract it from both sides by adding (16.0 m/s2) to both sides. (Similar to what I do to get the 0 m2/s2 over to the right side.) That effectively moves it from one side to the other and also changes its sign. Then, I’d be left with d, wouldn’t I? Do I have to consider the entire (-16.0 m/s2)d as one quantity? If I do that, what operation am I engaging in order to move the whole thing to the other side? What am I doing to both sides? Am I adding 16.0 m/s2d to both sides? Can I leave the d alone or do I have to consider it as part of the expression? Jesus, I should have paid attention to algebra.

#1) 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2)*d

#2 (16.0 m/s2)*d = 900 m2/s2 - 0 m2/s2

2

Um, I’m not quite awake yet, but it seems like all of your units are the same, right? So we have

  1. 0 = 900 - 16d
  2. 16d = 900

I think you did it right.

ETA: Crap, nevermind. I’ll give it another look and get back to you.

EETA: No, I think it still works. I’ll let someone who’s awake tell me if we’re both wrong.

3

(-16.0 m/s[sup]2[/sup]) isn’t added to anything, so you can’t just subtract it out (or rather, you could, but it’d just make things worse rather than better). The thing that’s added is the entire (-16.0 m/s[sup]2[/sup])*d, so that’s what you have to subtract from both sides (in other words, adding (16.0 m/s[sup]2[/sup])*d ). Once you’ve done that, you now need to separate the acceleration piece from the distance (I’m assuming that you’re solving for d), and since they’re multiplied together, you do that by dividing both sides by (16.0 m/s[sup]2[/sup]) .

BellBookCandle, you should always leave the units in the problem. It probably won’t get you in trouble in a simple problem like this, if you just put them back in at the end, but in a more complicated problem keeping them in throughout is usually the easiest way of figuring out what the end units are, and double-checking units will reveal most mistakes you might have made in the problem, so it’s a good check on your work.

4

Long time no algebra, but I think the operation is just called “re-arranging”, (swapping a complete term from one side to the other and changing the sign).

5

If you were to add 16.0 m/s2 to both sides of this, you’d have

0 m2/s2 + 16.0 m/s2 = 900 m2/s2 + (-16.0 m/s2)*d + 16.0 m/s2

or

0 m2/s2 + 16.0 m/s2 = 900 m2/s2 + 16.0 m/s2 + (-16.0 m/s2)*d

Because multiplication “comes before” addition in the order of operations, you cannot simplify the right side by adding together the 16.0 m/s2 + (-16.0 m/s2).

What are you trying to do? That is, which thing do you want to solve for?

6

I know, although it came up much more in science than in math. It was always easier to leave them in. Here though, it seemed like because of the formatting of the message board, it was really just cluttering up what seemed to be a pretty straightforward equation.

7

I guess I should have mentioned that :smack: - yes, I was solving for d.

8

Since the equation is linear (in d), the way to solve it is to isolate the d term (i.e. get it on one side by itself) and then divide both sides by the coefficient (the thing that d is multiplied by). This is the general strategy to use for solving linear equations. If there had been more than one d term, your first step would be to get all the d terms on one side and all the non-d terms on the other (by adding the same thing to both sides).

9

If you are solving 1) for d, then there is no reason to move the d term at all. Simply subtract 900 m[sup]2[/sup]/s[sup]2[/sup] from both sides of the equation, then divide through by -16 m/s[sup]2[/sup].

10

CC, I think the basic algebra fact you are reaching for is one you were never taught properly. When you are dealing with an equation, that means the things on each side of the = sign are the same, and what you do to one side you must do to the other. (Watch out for dividing by zero - it’s a big no-no!)

So for your equation 1:

0 m2/s2 = 900 m2/s2 + (-16.0 m/s2)*d

To solve for d - in other words to get d by itself on one side of the = sign - you can, as DSYoungEsq suggested subtract 900 m2/s2 from both sides, giving:

-900 m2/s2 = -(16.0 m/s2)*d

Then divide both sides by -16.0 m/s2, giving:

(-900 m2/s2)/(-16.0 m/s2) = d

From there you’ll need to divide both the numbers (-900/-16) and the units (m2/s2)/(m/s2).

Since you’re sticking a 2 next to a unit to indicate squaring the units come out to m.

Since you’re new at this, subtracting something from both sides was probably not your first thought, nor was dividing by a negative number probably a comfortable move for you.

That’s OK, you can have taken equation 1:

0 m2/s2 = 900 m2/s2 + (-16.0 m/s2)*d

and add +(16.0 m/s2)*d to both sides, giving:

(16.0 m/s2)*d = 900 m2/s2 + (-16.0 m/s2)*d + (16.0 m/s2)*d

Since you have a -x + x on the right hand side, your equation becomes:

(16.0 ms2)*d = 900 m2/s2

Then you divide both sides by the thing that multiplies d (the 16.0 m/s2), and you get

d = (900 m2/s2)/(16 m/s2)

That doesn’t look exactly like what I had above, until you remember that (-a/-b) = ([-1a]/[-1b]) and that [-1/-1] = 1, therefore (-a/-b) = (a/b).

FYI, you were using m2 to show “meters squared” and s2 to show “seconds squared”. Another way to indicate raising something to a power when you don’t have superscripts is to use the caret symbol ^, which is shift+6 on most keyboards. mm = m^2 . If you don’t have a caret, use **: mm = m**2 .

I hope that helps, I’ve only hd one cup of caffeine so far today.

11

Yep, that helps a lot. (Including your being nice enough to excuse me for my ignorance). I was trying to do the same thing to both sides. I got stuck on the part that had a quantity in parentheses multiplied by d. I wanted to get rid of that quantity and isolate d. And the example I was working from, which I listed as equations #1 and #2 showed the whole thing being moved over to the left side. And I thought I should be able to just move the quantity (-16 m/s^2), to leave the d on the right, by dividing both sides by (-16 m/s^2), which just got more complicated. I imagine, however, that such a move ought to also work, and I’ve just forgotten how to manipulate these basic parts of equations, the rules, the way that it all works. And logic wasn’t quite close enough for me to rely on. Tx.

12

CC, don’t be hard on yourself! You are admitedly rusty, and you may not have gotten the hang of it in the first place. Math and the sciences are often poorly taught!

Also, your problem was complicated enough that you may not have seen it was “just”

0 = b - (a*x)

with decorations (all those units and junk). With experience you can just glance at that and say “Of course! x = b/a, it’s obvious!” Without experience, you look at that and say “Bwah??” We all have to learn this stuff - nobody’s born with it.

13

Aha! Yes. When I look at it like that, I see that it really “just” means that b=a*x, since if they are added together they = 0. I do think that what I really need is more familiarity with it all. I’m trying to recapture it, as part of some basic physics, and after all these years, it’s hard to do on my own.

14

CC, this process is called rearranging or simplifying, but the technical process is called addition (or subtraction) of equations.

If a=b and c=d, then a-c=b-d. Essentially, you’re making it equation 1-equation 2. For your problem, you have a=b+cd. So you create a second equation, b=b. That’s obviously true. Now you subtract them, and get a-b=b+cd-b. That reduced is a-b=cd. That’s the essence of how you move the b term across to the other side and change it’s sign. The reason the c and d are glued together is because if you’d created a different formula (c=c) and subtracted it, you end up with a-c=b+cd-c. You can’t simplify that. You could, if you want, make the second expression b+c(d-1) for the same reason that 1000 five-dollar bills minus a five-dollar bill = 999 five-dollar bills. But that doesn’t help you solve anything.