Solving 3 Unknown Variables in Physics II

[Mirth]

So I'm taking Physics II, and I'm running into some basic algebra problems that are just going right past me, so I ask for assistance in brushing up my memory.

Here is an example problem, and it's answers, with 3 unknowns with 3 equations:

A + B = C

0.1A + 12 = 0.01B + 14

0.01B + 1.20C = 12


(Anwers: A = 19.1, B =-9, C=10.1)


I apologize profusely for my retardation, and any step by step guidance would be greatly appreciated.

 

[CompuChip]

The two common ways to solve these are:

Multiply all equations with suitable numbers, then add or subtract them to obtain a single equation for one of the variables

or

Use two of the equations to express all variables in one other and then substitute them (for example rewrite the second to B = ... and the third to C = ..., then plug into the first to get an equation for A).

Once you have found one of the values, you can discard on of your equations and solve the remaining system in the same way, until you find values for all of them.

A simple example with two equations in two unknowns:
2x + 3y = 6
x - 4y = 8.

The first way would be to multiply the second equation by 2 and then subtract them:
2x + 3y = 6
2x - 8y = 16
==> (2x - 2x) + (3y - -8y) = (6 - 16)
==> 11y = -10
==> y = -10/11

The second way is rewriting the second equation, for example
2x - 8y = 16 to x = (16 + 8y)/2 = 8 + 4y
then plugging it into the first one
2x + 3y = 6 with x = 4y + 8
==> 2(4y + 8) + 3y = (8y + 3y) + 16 = 6
==> 11y = 6 - 16
==> y = -10/11

Now you have one of the values, pick any of the two equations and back-substitute, e.g.
2x + 3y = 2x + 3(-10/11) = 6
==> 2x = 6 + 30/11
==> x = 3 + 15/11 = 48/11.

 

[Mirth]

Thank you a BUNCH, that cleared things up for me so much and I got the right answers.

Thanks!