Solving Complex Equations: Step-by-Step Guide for Large Numbers"

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In summary, In equation one, 48A - 32B - 2C = 0, the first equation is solved for C and it becomes 12B. Similarly, in equation two, -32A + 46B - 11C = 0, the first equation is solved for A and it becomes 13C. Finally, in equation three, -2A - 11B + 13C = 54, the first equation is solved for B and it becomes 12C.
  • #1
wellcoughed
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I've been looking on youtube to try and learn how to do these more complex equations but all the examples use small numbers that easily fall into place.

Ive got
Equation One: 48A - 32B - 2C = 0
Equation Two: -32A +46B - 11C = 0
Equation Three: -2A - 11B + 13C = 54

Ive mulitplied equation three by -16 which gives me:

NewThree: 32A + 176B - 208C = -864

Then I added equations Two and NewThree:
Equation Two: -32A + 46B - 11C = 0
NewThree: 32A + 176B - 208C = -864

Which canceled out the 32A and gave me:
222B -219C = -864

I then took Equation Three and mulitplied it by 24 which gave me:
Anothernewthree: -48A - 264B + 312C = 1296

Then added Equation One and Three together:

Equation One: 48A - 32B - 2C = 0
Anothernewthree: -48A - 264B + 312C = 1296

Which canceled out the 48A leaving me with:

-296B + 310C = 1296

Which left me with:
222B -219C = -864
-296B + 310C = 1296

And it's at this point that I get stuck :/

Is this the best route to take? Is there another method that is better to take for these larger numbers? The examples on youtube use examples that have small numbers and are easily solved, but I don't know what step to take next for this one :/

If anyone could help me I'ld be very grateful
 
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  • #2
wellcoughed said:
I've been looking on youtube to try and learn how to do these more complex equations but all the examples use small numbers that easily fall into place.

Ive got
Equation One: 48A - 32B - 2C = 0
Equation Two: -32A +46B - 11C = 0
Equation Three: -2A - 11B + 13C = 54

Ive mulitplied equation three by -16 which gives me:

NewThree: 32A + 176B - 208C = -864

Then I added equations Two and NewThree:
Equation Two: -32A + 46B - 11C = 0
NewThree: 32A + 176B - 208C = -864

Which canceled out the 32A and gave me:
222B -219C = -864

I then took Equation Three and mulitplied it by 24 which gave me:
Anothernewthree: -48A - 264B + 312C = 1296

Then added Equation One and Three together:

Equation One: 48A - 32B - 2C = 0
Anothernewthree: -48A - 264B + 312C = 1296

Which canceled out the 48A leaving me with:

-296B + 310C = 1296

Which left me with:
222B -219C = -864
-296B + 310C = 1296

And it's at this point that I get stuck :/

Is this the best route to take? Is there another method that is better to take for these larger numbers? The examples on youtube use examples that have small numbers and are easily solved, but I don't know what step to take next for this one :/

If anyone could help me I'ld be very grateful

Divide 222B -219C = -864 by 222 and -296B + 310C = 1296 by 296 and add them. It's nice to stick to the whole numbers for small numbers but you are going to get fractions in the end anyway. May as well start now.
 
  • #3
wellcoughed said:
I've been looking on youtube to try and learn how to do these more complex equations but all the examples use small numbers that easily fall into place.

Ive got
Equation One: 48A - 32B - 2C = 0
Equation Two: -32A +46B - 11C = 0
Equation Three: -2A - 11B + 13C = 54

Ive mulitplied equation three by -16 which gives me:

NewThree: 32A + 176B - 208C = -864

Then I added equations Two and NewThree:
Equation Two: -32A + 46B - 11C = 0
NewThree: 32A + 176B - 208C = -864

Which canceled out the 32A and gave me:
222B -219C = -864

I then took Equation Three and mulitplied it by 24 which gave me:
Anothernewthree: -48A - 264B + 312C = 1296

Then added Equation One and Three together:

Equation One: 48A - 32B - 2C = 0
Anothernewthree: -48A - 264B + 312C = 1296

Which canceled out the 48A leaving me with:

-296B + 310C = 1296

Which left me with:
222B -219C = -864
-296B + 310C = 1296

And it's at this point that I get stuck :/

Is this the best route to take? Is there another method that is better to take for these larger numbers? The examples on youtube use examples that have small numbers and are easily solved, but I don't know what step to take next for this one :/

If anyone could help me I'ld be very grateful

The standard way to solve such problem is also the simplest way: just solve successively for some variables, one-by-one, until you have just one variable left. Let me illustrate: your original equations are:
(1) 48A - 32B - 2C = 0
(2) -32A +46B - 11C = 0
(3) -2A - 11B + 13C = 54
Pick one of the equations and one of the variables; solve for that variable in terms of the others. It does not matter in the end which equation and variable you pick, as long as you do exact arithmetic. (It does matter when using a computer using floating-point computations---but that's another story for another time.)

So, let's pick eq (3) and solve for A in terms of B and C: (3) --> 2A = -54 -11B + 13C, so

(3a) A = -27 - (11/2)B +(13/2)C.
Substitute this expression for A in eqs (1) and (2). For example, eq(1) becomes 48[ -27 - (11/2)B +(13/2)C] = 32B - 2C = 0, or -1296 - 296B + 310C. Do the same type of operations on eq (2). Our new equations are:

(1a) -1296 - 296 B + 310 C = 0
(2a) 864 + 222 B - 219 C = 0

Now do the same to (1a)-(2a): pick an equation and a variable and solve. For example, let's solve for B from eq (1a): 296 B = -1296 + 310 C, or B = -(162/37) + (155/148) C.

Substitute this into (2a) to get -108 + (27/2) C = 0.
Our new system of equations is:

(1b) B = -(162/37) + (155/148) C
(2b) -108 + (27/2) C = 0
(3a) A = -27 - (11/2)B +(13/2)C

Now solve for C from (2b) to get C = 8. Substitute that into (1b) to get B = 4. Substitute those two values into (3a) to get A = 3.

It is annoying to carry along all those fractions, but it is usually unavoidable. Keeping the exact fractions allows us to obtain an exact final solution; if we rounded off decimals we would be making some errors that could result in some errors in the final solution.
 
  • #4
I'ld just like to say many thanks to those who were kind enough to take time out of their day to help me.

You guys have made the world a little bit brighter for me :)

Thanks again
 
  • #5
It hurts my eyes to see your first equation, which has a factor of 2 that can be cancelled. After doing this, you immediately have C = 24A - 16B. If you substitute this into the second equation to eliminate C, you get 111B = 148A. If you reduce this to lowest terms, you quickly get A = 3B/4. Substituting this back for C gives you C = 2B. Those zeros on the right hand sides of equations 1 and 2 should have driven you to this approach. Finally, you can substitute for A and C in the third equation to solve explicitly for B.
 
  • #6
Yeah I threw away the method I used in my original post.

I haven't done it the same way as you guys, (I got values for A,B,C after about 9 steps while you guys seemed to have done it in 5 or 6)

However battling the logic out for myself has helped me understand the process better, and I did think about going back and simplifying my method but considering it's the first time I've had to do one (without any tuition on how to do 3 variable equations from my lecturer) so I'm going to keep my 9 step original :)

Thanks again everyone :)
 

FAQ: Solving Complex Equations: Step-by-Step Guide for Large Numbers"

What is a 3 variable equation?

A 3 variable equation is an algebraic equation that contains three variables, or unknown quantities, and mathematical operations such as addition, subtraction, multiplication, and division. These equations are used to solve for the value of the variables and find the relationship between them.

How do you solve a 3 variable equation?

To solve a 3 variable equation, you need to have three equations with three variables. You can use substitution or elimination methods to find the value of each variable. By substituting the value of one variable into another equation, you can solve for the other variables. In elimination, you can eliminate one variable by adding or subtracting two equations together and then solve for the remaining variables.

What are some real-life applications of 3 variable equations?

3 variable equations have many real-life applications, such as in physics, chemistry, economics, and engineering. For example, they can be used to model the relationship between temperature, pressure, and volume in a gas, or the supply, demand, and price of a product in economics. They are also used in optimization problems, such as finding the maximum or minimum value of a function with multiple variables.

Can a 3 variable equation have more than one solution?

Yes, a 3 variable equation can have infinite solutions. This is because there are three variables, but only two equations. Therefore, there are an infinite number of combinations of values that can satisfy the equations. However, if there are more than three equations, the system can have a unique solution.

What is the difference between a 2 variable and 3 variable equation?

The main difference between a 2 variable and 3 variable equation is the number of unknown quantities. A 2 variable equation contains two variables, while a 3 variable equation contains three variables. This means that a 3 variable equation is more complex and requires more equations to solve. Additionally, a 3 variable equation can have a unique solution, infinite solutions, or no solution, while a 2 variable equation can only have a unique solution or no solution.

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