Is it possible to do other kinds of systems of linear equations?

[batx27]

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

[MarkB]

(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")

[batx27]

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

[Aries]

Assuming the parameter k>0, you should see something like:
x=14/(k-20) and y=(75-2k)/(k-20)
Best,

Aries ;-)

[nickapos]

(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.