We generally think of linear equations in the form , where is the gradient and is the -intercept. There are, however, other ways including: , , and (which is line-slope form). All of these are functions for finding straight lines, and should be middle-school knowledge. In all examples, we input an value, and get a value as an output, and this correlation is linear (hence, equation).

When given cartesian coordinates and , we're able to get the gradient, , using the following formula: , which takes the and divides it by . We colloquially say it as "change in by change in ", which happens to be the definition of the gradient!

We can get either: , 1, or 0 solutions. The number of solutions when dealing with two linear equations abides by these rules:

In simultaneous equations, these solutions represent where two equations intersect, and in intercepts, where equation(s) intercept the or -axis.

Now that we've got the basics of line equations, we can apply them to find out things like perpendiculars, and perpendicular bisectors.

When forming the perpendicular bisector of a line, we take the negative reciprocal of the gradient, which is defined as .

When faced with a question like "Find the perpendicular bisector of with the endpoints at and ", we need to note that the perpendicular is a line (perpendicular to ) created on the midpoint between two points and .

To find the midpoint, we use , which is the average (mean, if you would like to be specific) of the distances of and . This logic can also be applied to finding the centroid of a triangle, which is found by , however we don't have to worry about that for now.

If we want to find the perpendicular bisector (line perpendicular to at a midpoint of a segment), we can just insert the reciprocal of the gradient () and the coordinates of the midpoint into the following equation: , where are the midpoint coordinates, is the gradient, and form the function itself.

When solving simultaneous equations, especially in IB, the general elimination method isn"t enough. Gaussian elimination using augmented matrices provides a more efficient approach. This involves representing equations as matrices and applying row operations to solve. Don't worry about understanding matrices fully now; they'll be introduced in a later chapter.

In order to solve them using matrices, we need to turn the formed matrix into what's called "reduced row form" (don't worry, you don't have to memorize that word). This involves us getting the following forms of a matrix:

Note that in exams, you might not explicitly be asked 'give the value of that produces a single solution'. Instead, exams might say: "consistent systems" (systems have at least one solution), "inconsistent systems" (systems have no solutions), and "unique solution" (systems have exactly one solution), and then the obvious "infinite solutions" (equations are linearly dependent of each other [hint hint], meaning there's an infinite number of solutions).

We can input equations into a matrix, and manipulate the contents of the matrix to get our required solutions. An example of said manipulation is below:

Having a solution to two sets of equations is nice, however we can get infinite solutions if they never intersect (remember Note N01.0a from the top). With this, you have to "introduce" a variable yourself, and express the answer as that variable. For now, you won't have to worry about this. I found it quite hard to find questions that worked to produce infinite solutions, and always found ones that produce . Think about why that might be!

Try some yourself. Remember, there are different reduced-row-matrix forms that could be achieved!

Common examples of applications involves establishing relationships (in physics or chemistry) and finding break-even points (in finance). Generally because they're linear equations, the applications are quite easy to solve, so if you just want to glance over this, you don't have to solve them if you don't want to.