Inequations with two variables

Once we can deal with first degree inequations with a variable (which we call $$x$$), we will introduce a second variable that we will call $$y$$. Therefore, now we have algebraic expressions formed by numbers, $$x$$, $$y$$ and a symbol of inequality.

$$$ x+y < 1$$$

When we are solving an inequation, we say that we have solved it when we obtain a result like $$x < a$$. As expected, now we will not be able to do the same because we have two variables.

So: How do we express the result? What does it mean to solve an inequation of two variables? Next, we will answer these questions.

Let's start with an example and later we will give a general indication about how to solve these inequations.

As we can see, the expression $$ x+y < 1$$ is similar to $$ x+y = 1$$, and that the last one is no more than a straight line that can be represented in the plane, by reason of $$ x+y = 1 \Rightarrow y=-x+1$$. This means that the points that satisfy the equality $$x+y = 1$$ are exactly the points of the straight line.

Then, we can say that the points of the straight line do not satisfy the inequality given by the inequation $$ x+y < 1$$, since these points are such that $$x + y$$ is one and we are looking for points where $$x + y$$ is less than one.

However, we can express our inequation as: $$ x+y < 1 \Rightarrow y < -x+1$$ and the previous straight line divides the plane exactly in two parts, one over the straight line and one below. Then, as our inequation is $$y < -x+1$$, the points that satisfy the inequation are exactly those below the straight line.

Resolution algorithm:

Let's take a look at how we have solved this inequation of two variables (the previous example):

  • We have separated the variables, one on each side of the inequation, usually leaving the $$y$$ alone on one side, and the $$x$$ on the other with the free coefficients, expressing it as $$y < ax+b$$ or $$y > ax+b$$.

  • We have drawn the line induced by the inequation (the straight line $$y = ax+b$$).

  • We have chosen an area of the plane: the one that is above or below depends on our inequation. We have made the following choice:

    • If $$y < ax+b$$, then we say that it is the area below the straight line.
    • If $$y > ax+b$$, then we say that it is the area above the straight line.
  • This area of the plane is the one that is a solution to our inequation.

We have to bear in mind that our inequation might have an inequality of the type $$ < $$, $$ > $$ $$\leqslant$$ or $$\geqslant$$. In the first two cases, the area in the plane that we choose does not contain the points of the straight line, while in the last two cases, when we have "less than or equal to" or "greater than or equal to", we are taking the points of the straight line in addition to the area above or below in the plane , as a solution for our inequation.

Then: does giving a region in the plane amount to giving a solution of an inequation? Yes, but we must think that it is the same to give this region in the plane as it is to give the inequation where the $$y$$ is on one side and the $$x$$ and the free coefficients on another. So we can say that solving an inequation is giving an expression of the type $$y < ax+b$$ or $$y > ax+b$$ and saying that the points we take are below or above the line $$y = ax+b$$ and, finally, if we take those points of the line or not.