How to solve linear equations in one variable

a repository of mathematical know-how

How to solve linear equations in one variable

Title: *

Area of mathematics: *

A comma-separated list of areas of mathematics to which this article applies. Use ">" to tag in a subcategory. Example: Analysis > Harmonic analysis, Combinatorics

Keywords:

A comma-separated list of keywords associated with this article. Example: free group

Used in:

A comma-separated list of examples of where this technique is used. Example: Cauchy-Schwarz inequality

Parent articles:		Order

Body:

[QUICK DESCRIPTION]

Linear equations in one variable, also known as first-degree equations, are the simplest kind of equations. A typical example is the equation $3x+2=11.$ More generally, we will be concerned with any equation of the form $ax+b=c,$ where $a$, $b$, and $c$ are known numbers and we are trying to find $x.$ One can often solve such equations by trial and error, but there is also a straightforward method of solving them systematically. We will also describe the ''Egyptian method'', which can be thought of as a simple instance of the Newton's method.

=== Method 1: Simple trial and error ===

First, however, we discuss a different approach. Suppose that our equation is $7x+2=51.$ We can try some numbers, adjust them, and see if we manage to hit the right answer. For example, trying $3$ as our $x$ we get $7*3+2=23$ and with $5$ we get $7*5+2=37.$ As we increase our $x,$ it seems that we increase our result, so if we want to get $51$ then we will want to try an even higher $x.$ If we try a considerably higher $x$ such as $8,$ then we get $7*8+2=58,$ which has passed the mark a little. Coming down to $7,$ we find that it does the trick, since $7*7+2=51$.

But what happens if the equation is instead $7x+2=54,$ say? From the previous calculations we see that our $x$ must be between $7$ and $8,$ so it cannot be a whole number. We could try decimals such as $7.5,$ but then our trials and errors start to take longer and longer, and in fact no terminating decimal will give the right answer: it will only ever  approximate the exact solution.

[GENERAL DISCUSSION]

What can we do about this problem? Well, in any kind of equation we are trying to find an object that satisfies some equality. In our case we are trying to find a number "x" that makes two expressions such as $7x+2$ and $37$ equal to each other. The main idea that underlies the usual approach to this problem is that ''if'' the two expressions really do give the same number, then if we do the same thing to both of them (such as, say, subtracting $2$), then they will ''still'' give the same number. If we can somehow manage to transform one expression into $x$ and the other into a number, then we have shown that if $x$ is a solution, then it has to be that number. Let us see how this can be turned into a systematic method for solving linear equations in one variable.

=== Method 2: Operating on both sides of the equation ===

Let us take the equation $3x+11=25.$ As just noted, we can add to or subtract from both sides any number we want and the equality will be maintained. For instance, if $3x+11=25,$ then $3x+11+7=25+7,$ which allows us to say that $3x+18=32.$ However, this equation doesn't seem any easier to solve than the original one. However, there is one way in which the left-hand side can be simplified: if we subtract $11$ from it. So let us subtract $11$ from both sides. Now we get $3x+11-11=25-11,$ which tells us that $3x=14.$ This looks like an easier equation, but if we add or subtract anything then we will not simplify it further, so if we want to use the general idea of doing the same to both sides, then we will have to do something else. What are we trying to simplify? Well, we were hoping to get $x$ on the left-hand side, and we can do that easily if we turn to division: to get rid of the $3,$ all we have to do is divide by it. So let us divide by $3$ on both sides: we get $\frac{3x}{3}=\frac{14}{3},$ which tells us that $x=\frac{14}{3}$.

This method is both practical and failsafe: an equation of the form $ax+b=c$ will have the solution $x=\frac{c-b}{a}$.

=== Method 3: Egyptian Method ===

Again, we illustrate the method with an example. Let us take the equation $3x+11=32.$ This method starts with some trial: let us take $x=2.$ This gives us $3*2+11=17,$ but to get to $32$ we are still missing $32-17=15.$ Since our $x$ is multiplied by $3,$ every $1$ that we add to the $x$ will give $3$ more to the total. Therefore, to get the extra $15$ we will need to add $15/3=5$ to our $x=2.$ And indeed, if we take $x=2+5=7$ then we get $3*7+11=32$.

This method always works too and in fact is an instance of [[w:Newton's Method]] for approximating solutions of more general equations. In our case this method always gives the correct answer in the first step. Newton's method usually continues indefinitely, or until one gets a good enough approximation to a solution, which is often the best we can hope for anyway.

This is a stub

A stub is an article that is not sufficiently complete to be interesting.

Notifications

File attachments

Changes made to the attachments are not permanent until you save this post. The first "listed" file will be included in RSS feeds.

Attach new file:

Images are larger than 640x480 will be resized. The maximum upload size is 1 MB. Only files with the following extensions may be uploaded: jpg jpeg gif png svg.

Revision information

Log message:

An explanation of the additions or updates being made to help other authors understand your motivations.

How to solve linear equations in one variable

Recent articles

Active forum topics

Recent comments