Question on solving system of equations with constants....

In summary, solving a system of equations with numerical constants in front of each term can be done using Gaussian elimination, but when constants are represented by symbols, substitution is typically used. However, for a 3 x 3 system, Cramer's Rule can also be used but may become computationally difficult. Elimination can also be used for systems with symbols, as demonstrated in the conversation.
  • #1
datahead8888
10
0
It's fairly straight forward to find information on how to solve a system of equations like this:

2x + 3y + 4z = 1
3x + 4y + 3z = 2
4x + 5y + 3z = 3

It has numerical constants in front of each term. You could use Gaussian elimination and solve for one, infinite, or no solutions. (The above example is completely random). It works because you can cancel terms out.

It's a bit less straightforward to find information on how to solve it when you have constants represented by symbols:
ua*x + ub*y + uc*z = 1
va*x + vb*y + vy*z = 2
x + y + z = 3

Here we want to solve for x, y, and z, each expressed in terms of ua/ub/uc/va/vb/vc only. You can't use elimination here because terms such as ua and va will not cancel out. (This example is slightly random too - I just need to learn the technique).

The only method I'm aware of is substitution, though it seems to spill onto multiple pages (the above system is simpler than what I need to solve). Is there any easier way to attack something like this? Any links to online references would be much appreciated.
 
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  • #2
For a 3 x 3 system such as you have here, Cramer's Rule isn't too bad. Because it's all determinants, though, it gets computationally unwieldy very quickly.
 
  • #3
Ackbach said:
For a 3 x 3 system such as you have here, Cramer's Rule isn't too bad. Because it's all determinants, though, it gets computationally unwieldy very quickly.

That looks extremely helpful. It probably still requires mangling, but it least it gives a lot of order to the mangling. I especially like how wikipedia already has standard forms posted.

I'll post back if I have issues - thanks a lot!
 
  • #4
I'm not so sure elimination doesn't work here. Assuming you meant subscripts $u_{a}$, etc., we have:

$$
\left[ \begin{array}{ccc|c} u_a &u_b &u_c &1 \\ v_a &v_b &v_c &2 \\ 1 &1 &1 &3 \end{array}\right]_{ \begin{align*}R_{1} & \leftrightarrow R_{3} \\ R_3 & \leftrightarrow R_2 \end{align*}} \to $$
$$ \left[\begin{array}{ccc|c}1 &1 &1 &3 \\ u_a &u_b &u_c &1 \\ v_a &v_b &v_c &2 \end{array}\right]_{ \begin{align*}-u_a R_1+R_2 &\mapsto R_2 \\ -v_a R_1 + R_3 &\mapsto R_3 \end{align*}} \to $$
$$\left[\begin{array}{ccc|c}1 &1 &1 &3 \\ 0 &u_b-u_a &u_c-u_a &1-3u_a \\ 0 &v_b-v_a &v_c-v_a &2-3v_a \end{array}\right]_{R_2/(u_b-u_a) \mapsto R_2} \to
\left[\begin{array}{ccc|c}1 &1 &1 &3 \\ 0 &1 &\frac{u_c-u_a}{u_b-u_a} &\frac{1-3u_a}{u_b-u_a} \\ 0 &v_b-v_a &v_c-v_a &2-3v_a \end{array}\right], \quad \text{for } u_b-u_a\not=0.
$$
Continuing on with the row operation $-(v_b-v_a)R_2+R_3 \mapsto R_3$ yields
$$\left[\begin{array}{ccc|c}1 &1 &1 &3 \\ 0 &1 &\frac{u_c-u_a}{u_b-u_a} &\frac{1-3u_a}{u_b-u_a} \\ 0 &0 &\frac{(v_a-v_b)(u_c-u_a)}{u_b-u_a}+v_c-v_a &\frac{(1-3u_a)(v_a-v_b)}{u_b-u_a}+2-3v_a \end{array}\right].$$
Now you can do back substitution to finish.
 
  • #5


Thank you for your question. Solving systems of equations with constants can be done using various methods such as substitution, elimination, or matrix manipulation. In cases where the constants are represented by symbols, substitution is usually the most efficient method.

One way to approach the problem is to first solve for one of the variables in terms of the others. For example, in your system of equations, you can solve for z in terms of x and y by rearranging the third equation to be z = 3 - x - y. Then, you can substitute this expression for z into the first two equations, giving you two equations with two variables (x and y) that can be solved using substitution or elimination.

Alternatively, you can also use matrix manipulation to solve systems of equations with constants. This involves setting up a matrix with the coefficients of the variables and constants, and then using row operations to reduce the matrix to a triangular form. From there, you can solve for the variables using back substitution.

I would recommend searching for resources on "solving systems of equations with constants" or "solving systems of equations with symbols" to find more detailed explanations and examples. Online references such as Khan Academy, MathIsFun, and MathHelp.com also have step-by-step guides and practice problems for solving systems of equations with constants.

I hope this helps. Good luck with your problem!
 

Related to Question on solving system of equations with constants....

1. What is a system of equations with constants?

A system of equations with constants is a set of equations that contains both variables and constants. These equations can be solved simultaneously to find the values of the variables that satisfy all the equations in the system.

2. How do I solve a system of equations with constants?

To solve a system of equations with constants, you can use various methods such as substitution, elimination, or graphing. The chosen method will depend on the type of equations and the number of variables in the system.

3. Can a system of equations with constants have more than one solution?

Yes, a system of equations with constants can have one, infinite, or no solutions. This will depend on the relationships between the equations in the system.

4. What is the importance of solving a system of equations with constants?

Solving a system of equations with constants is important in various fields, such as science, engineering, and economics. It allows us to find the values of variables that satisfy all the equations and can help us make predictions and solve real-world problems.

5. Are there any tips for solving a system of equations with constants?

Some tips for solving a system of equations with constants include: 1) Choose the most efficient method based on the type of equations, 2) Simplify the equations by combining like terms, 3) Substitute or eliminate variables to reduce the number of equations, and 4) Check the solution by plugging in the values to all the equations in the system.

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