Jtdylktuy

I have an equation that has 3 unknown values ($x_1$, $x_2$ and $x_3$)
I know how to solve equations when only 1 $x$ exist but how do you solve an equation with 3 $x$ like this one?

Thank you!

$$0.43125x_1 + 0.35x_2 + 0.21875x_3 = 100$$

"but how do you solve an equation with 3 x like this one?"

You don't. There isn't a unique solution, if that's what you want. In fact, the set of solution is an affine space of dimension $2$.

I understand. I perheps put the problem incorrectly. I can try to describe the problem as it is and perheps there is another approach to it:

If we assume we have 100 occurences and our goal is to get the TOTAL number to
be 100.

All the numbers that we added should be equal to 100 but
I am not sure how to do this?

(The 3 different numbers should be equally balanced in the same way as the % of the occurences (43.125%, 35%, 21.875%))

I have tried with a solution but this doesn't seem correct.

1. In 43.125% of the occurences. I will add 43.125

2. In 35% of the occurences. I will add 35

3. In 21.875% of the occurences. I will add 21.875

Mathematical calculation: (100 occurences)

43.125 occurences * 43.125 = 1859.77

35 occurences * 35 = 1225

21.875 occurences * 21.875 = 478.52

1859.77 + 1225 + 478.52 = 3563.29

3563.29 / 100 occurences = 35.63

I get 35.63 here. The goal would be to get 100 here.
I wonder what I am doing wrong or how to set this up?

Federico is absolutely right, and the link he provides is pretty explanatory.

This is basic linear algebra so if you find this interesting you should review simultaneous linear equations and matrices, here’s a link.

Now, I’ll start by simplifying the question a little bit so it’s easier to follow and maybe you’ll understand it a little better, because at first the whole ‘no unique solution’ thing can be a little confusing.

Say we have the following relationship:

$$
0.4x + 0.3y + 0.2z = 100
$$

1. Solve for $x$:

$$
begin{align}
0.4x + 0.3y + 0.2z &= 100 \
0.4x &= 100 -0.3y -0.2z \
x. &= frac{100}{0.4} -frac{0.3}{0.4}y -frac{0.2}{0.4}z \
x &= 250 - 0.75y - 0.5z \
end{align}
$$

Now we have a solution for $x$ observe that $y$ and $z$ are independent variables, so set them to some arbitrary values

solve $y$

Let $y=s$ such that $s$ is some arbitrary real number

solve $z$

Let $z=t$ such that $t$ is some arbitrary real number

Conclusion

Thus the values for the variables must be :

$$
begin{align}
x &= 250 - 0.75s - 0.5t \
y &= s \
z &= t \
end{align}
$$

Any values satisfying that will hence satisfy the initial equation.

Example

Say we choose arbitrary values for $s$ and $t$ :

$$
begin{align}
s &= 4 \
t &= 2 \
end{align}
$$

Now by substitution we can solve for fixed equation values:

$$
begin{align}
x &= 250 - 0.75times 4 - 0.5 times 2 \
&= 250 - 3 - 1 \
&= 246 \
y &= 4 \
z &= 2 \
end{align}
$$

Now let’s test our values by substitution

$$
begin{align}
0.4x + 0.3y + 0.2z &= 100 \
0.4 times 246 + 0.3 times 4 + 0.2 times 2 &= 100 \
98.4 + 1.2 + 0.4 &= 100 \
99.6+ 0.4 &= 100 \
100 &= 100 \
end{align}
$$

So the solution is verified as working, but take in mind what Federico said, there are infinite solutions because the latter variables can be adjusted around any given $x_1$ value.

Thank you very much for that well explanation.

"there are infinite solutions because the latter variables can be adjusted around any given x1 value".

That seems to be true for me also as I have discovered that as well which was one of the questionmarks that I was start thinking of. I beleive a good solution then would be to test out the 2 second variables to "search" for a valid solution that is looked for.



Thank you very much for your help!