# Is this a known method for solving linear equations in 1 variable?

I thought of a method to solve linear equations in one variable and I want to know if this is a well-known method or not. It's an elementary method, which makes me think it has probably been known for a thousand years, yet I can't find any information about it online.

Given a linear equation with one variable, you test $$x=0$$ and $$x=1$$ as potential solutions, and if they are not, the differences between left and right sides of the equation can be used to determine the solution.

Denote the error caused by $$x=0$$ as $$E_{0}$$ and the error caused by $$x=1$$ as $$E_{1}$$. Then the solution to the equation is $$\frac{E_{0}}{E_{0}-E_{1}}$$

For example: Consider the equation $$5(6x-3)+1=2(7-9x)+4$$.

$$x$$ LHS RHS Error
$$0$$ $$-14$$ $$18$$ $$32$$
$$1$$ $$16$$ $$0$$ $$-16$$

The solution is $$x=\frac{32}{32-(-16)}=\frac{2}{3}$$.

This works by considering $$(0,32)$$ and $$(1,-16)$$ as points on a line, and then finding the $$x$$ intercept of that line.

This method works for special cases as well:

• The equation is true for any number if and only if both errors are $$0$$
• The equation has no solutions if and only if the errors are equal (to something other than $$0$$)

• Yes, as you point out. You may as well move everything to one side so you're looking for an $x$-intercept only. Then once you know two points, you know the line, so you know the $x$-intercept. Dec 14, 2021 at 3:17
• Some context: I'm teaching algebra and I was trying to think of a way to solve these types of equations without simplifying expressions or using inverse operations. I'm just wondering if this particular method has been discussed or written about previously. Dec 14, 2021 at 3:20
• BTW, did you know that there is a separate Stackexchange for History of Science and Math? To get answers about previous versions of your method, you might want to ask over there. Dec 14, 2021 at 18:57
• This is known as double false position or regula falsi. Dec 14, 2021 at 20:00
• Thank you @franzlemmermeyer. You provided exactly what I was looking for. Wikipedia says a version of this method was in the Rhind Papyrus, so my guess of "a thousand years old" was a serious underestimate! Dec 15, 2021 at 2:02

I'm pretty sure it's equivalent to some of the more standard methods of solution, but I think it's pretty cool that you came across it on your own.

If I were trying to understand what was going on, I would say that an arbitrary equation of your form can be written as $$f(x) = g(x)$$ where $$f$$ and $$g$$ are both linear. Then you are transforming it to $$f(x) - g(x) = 0$$ where $$f(x) - g(x)$$ has the form $$ax+b$$. Then your "errors" are given by $$E_0 = f(0) - g(0)$$, and $$E_1 = f(1) - g(1)$$. Take a look at how $$E_0$$ and $$E_1$$ are related to $$a$$ and $$b$$, and you might get a bit more insight.

BTW, another way of looking at your solution is that $$E_0$$ expresses how "wrong" you are by guessing $$0$$ for $$x$$ (otherwise knows as the $$y$$-intercept). And $$E_1 - E_0$$ tells you how much you "change the error" when you increase $$x$$ by $$1$$. So dividing your "error" by that tells you how much you need to change $$x$$ (from $$0$$) to make your "error" disappear. Notice that this also predicts that you don't have to start at $$0$$, and you could come up with other formulas like $$7 - E_7/(E_8 - E_7)$$ or $$4 - \frac{E_4}{(E_6 - E_4)/2}$$ where in the second one we compare the "errors" at points that are $$2$$ apart, so to get the "change in error per unit change in $$x$$" we have to divide by $$2$$.

You can also describe your two special cases in these terms. If the errors are the same at two different values of $$x$$, then you've "got no gas" - changing the value of $$x$$ doesn't change the error.

So if you had solutions for those values of $$x$$, changing $$x$$ doesn't change the errors, so you can change $$x$$ to whatever you want and you still have a solution. And if you didn't have solutions for those two values of $$x$$, change $$x$$ as much as you want and you'll never change the "error", so you can never get solutions.

(The more I use it, the more I like this way of thinking about solving these. Thanks for sharing.)

• Right. With $h(x)=f(x)-g(x)$, you're just giving $h(1)$ and $h(0)$. Work out the $x$-intercept of a generic line in terms of the coefficients and you'll see what's happening. Dec 14, 2021 at 3:28