How to Rearrange Equations: Solving for Unknown Variables

[Eisely]

Homework Statement

I have a page full of questions for my physics class and I've forgotten the basic steps to rearranging formulas. Here are 2 examples of the types of questions I have on my sheet.

Homework Equations

Solve the equation y=mx+b for m.

Solve the equation F=GMm/r^2 for m.

The Attempt at a Solution

I tried to remember what to do, and I know it has something to do with doing what you do to one side to the other. So for the first one I just put -x-b+y=m and I am pretty sure it's terribly wrong. For the second i played around and tried m=GM/Fr^2 which I'm also sure is pretty wrong.

I can't find a topic in my text to tell me how to do these operations as its expected that I already know them. IS there an online tool with these rules anywhere? Can you help me figure these two out? I think knowing these I can do the rest.

 

[berkeman]

Homework Statement

I have a page full of questions for my physics class and I've forgotten the basic steps to rearranging formulas. Here are 2 examples of the types of questions I have on my sheet.

Homework Equations

Solve the equation y=mx+b for m.

Solve the equation F=GMm/r^2 for m.

The Attempt at a Solution

I tried to remember what to do, and I know it has something to do with doing what you do to one side to the other. So for the first one I just put -x-b+y=m and I am pretty sure it's terribly wrong. For the second i played around and tried m=GM/Fr^2 which I'm also sure is pretty wrong.

I can't find a topic in my text to tell me how to do these operations as its expected that I already know them. IS there an online tool with these rules anywhere? Can you help me figure these two out? I think knowing these I can do the rest.

The main trick is to do the same thing to both sides of the equation, so that the = sign still holds true. Like if I add 1 to both sides of an equation, the = sign still holds true, right?

So you can multiply both sides by the same thing, or divide both sides by the same thing, and the = sign still holds true.

So if you have a = b x, and want to solve for x, what do you divide both sides by to isolate x? Remember that any ratio of a thing to itself is 1, so Z/Z = 1.

 

[berkeman]

And here's a wikipedia link for more info on basic algebra:

http://en.wikipedia.org/wiki/Algebra

.

 

[Eisely]

Okay, so I would divide both sides by x+b to isolate m, meaning m=y/x+b? and the second would be m=Fr^2/GM?

 

[eumyang]

Okay, so I would divide both sides by x+b to isolate m, meaning m=y/x+b?

No, that's wrong. Only the x is multiplied by m. The b is added on afterwards. You've got two operations here, so you'll need two steps to isolate the m.

For a hint on one of these steps, note that if
a + b = c,
you can subtract b from both sides and get
a = c - b.

 

[Rasalhague]

"Solve the equation y=mx+b for m."

"Okay, so I would divide both sides by x+b to isolate m, meaning m=y/x+b?"

Dividing both sides by x+b means: y/(x+b) = (mx+b)/(x+b). If the equation had been y = m(x+b) = mx + mb, then you could have isolated m by dividing by (x+b).

Some basic axioms and definitions. For any number, a, the number -a is defined as what you have to add a to get 0, thus a+(-a) = a-a = 0. And 1/a is what you have to multiply a by to get 1, thus a(1/a) = a/a = 1. The notation a² means aa, and a²/a² = (aa)/(aa) = (a/a) times (a/a) = 1 times 1 = 1. Multiplication "distributes" over addition according to the rule a(b+c) = ab+ac, and (b+c)a = ba+ca.

The other thing to remember is that what you do to (the whole of) one side of an equation, you should do to (the whole of) the other side, to be sure that the equation is still true.

A little practice playing around with rearranging equations and you'll soon get the hang of it again!

The Khan academy has some great videos. The Arithmetic series covers the basics at a gentle level ( http://www.khanacademy.org/ ).

 

[Eisely]

Oh! I think I get it.

so steps would be


y=mx+b

y-b=mx

y-b/x=m


m=y-b/x


?


I get why x+b isn't 'one term' now.

 

[Rasalhague]

y=mx+b

y-b=mx

You got the first step right.

y-b/x=m


m=y-b/x


?

Not quite. Here you divided the (whole of the) right side by x, so you should have divided the (whole of the) left side by x too, but instead you only divided one of its terms by x.

You're nearly there though... Dividing by x means the same as multiplying by 1/x. So (p+q)/r = 1/r times (p+q), and multiplication "distributes" over addition, so: (p+q)/r = p/r + q/r, which won't always be equal to p+q/r. This works just the same if p or q is negative: (p-q)/r = p/r - q/r, not necessarily equal to p - q/r.

By the way, if you want to test your answers:

http://www.quickmath.com/webMathematica3/quickmath/page.jsp?s1=equations&s2=solve&s3=basic

 

[Eisely]

Yeah, I wasn't clear in my answer but i had y-b all over x.

That website where I could test my answers really helped,

I did all the other questions, tested them and got them all right!

I think I'm finally figuring this stuff out ^.^!

 

[Rasalhague]

Yeah, I wasn't clear in my answer but i had y-b all over x.

To avoid that ambiguity, the rule is that if you want to multiply more than one term by some number, you have to use brackets, (p + q)r, or write it out in full as pr + qr. And the same goes for multiplying by 1/r, in other words dividing by r. Thus, (p + q)/r = p/r + q/r. Or if you can write them on two levels:

$$\frac{p+q}{r}=\frac{p}{r}+\frac{q}{r}$$

$$=(p+q)/r = \frac{1}{r}\left ( p+q \right )=r^{-1}(p+q)$$

$$\neq p + \frac{q}{r}$$

That website where I could test my answers really helped,

I did all the other questions, tested them and got them all right!

I think I'm finally figuring this stuff out ^.^!

Excellent!