Is it possible to do other kinds of systems of linear equations?
08-01-2023, 05:38 PM
Post: #1
 batx27 Junior Member Posts: 17 Joined: Jul 2023
Is it possible to do other kinds of systems of linear equations?
The linear solver only takes x+y=z
x+y=z
Is it possible to do an equation like
10x-4y=8
8y=kx-30
08-02-2023, 07:38 AM
Post: #2
 MarkB Junior Member Posts: 12 Joined: Aug 2021
RE: Is it possible to do other kinds of systems of linear equations?
(08-01-2023 05:38 PM)batx27 Wrote:  Is it possible to do an equation like
10x-4y=8
8y=kx-30

As the User Guide says (Section 16 on page 312 in my copy) equations must be in the form:
ax + by = k (for the 2x2 solver)
ax + by + cz = k (for the 3x3 solver)

Simply rearrange any equations to fit the appropriate form and then enter the coefficients into the boxes.

10x - 4y = 8 is already [10]x + [-4]y = [8]
8y = kx - 30 becomes [k]x + [-8]y = [30]
(presumably you know "k")
08-02-2023, 10:06 PM
Post: #3
 batx27 Junior Member Posts: 17 Joined: Jul 2023
RE: Is it possible to do other kinds of systems of linear equations?
Hey Mark. I actually don't know the value for K, so I wondering if there was a shortcut or program to find K, but I guess I will keep looking. Thanks tho
08-03-2023, 05:00 AM
Post: #4
 Aries Member Posts: 159 Joined: Oct 2014
RE: Is it possible to do other kinds of systems of linear equations?
Assuming the parameter k>0, you should see something like:
x=14/(k-20) and y=(75-2k)/(k-20)
Best,

Aries ;-)
08-03-2023, 05:51 AM
Post: #5
 nickapos Member Posts: 170 Joined: Sep 2021
RE: Is it possible to do other kinds of systems of linear equations?
(08-02-2023 10:06 PM)batx27 Wrote:  Hey Mark. I actually don't know the value for K, so I wondering if there was a shortcut or program to find K, but I guess I will keep looking. Thanks tho

There is no way to solve a 2 equation system with 3 unknowns. One of them (k in this case) will have to remain unknown and you will have to express the x and y in terms of k.
Your system will have infinite solutions, unless you have a restriction for k.
For each separate value of k you will have a different solution.
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