# Having a difficult time figuring out if linear equations is true or false.

I'm having a very difficult time figuring out if these linear equations are true or false: $$x + 2y > 6\\ x - y < 3$$ How do you identify if these equations are true or false? At first I thought I was supposed to write out the solution, then once I get the answer I was supposed to say if it was true or false. Is that right? I created this graph identifying both inequalies, but I just don't know how to figure it's true or not. If anyone could help that would be great.

• A solution is precisely that which makes the linear inequalities hold true. – Calvin Khor May 13 '15 at 12:39
• By checking false or true do you mean to verify if a point plugged in those equation to check if it satisfy it or not? – Mann May 13 '15 at 12:41
• @mann Yes! I'm supposed to graph my answe after. – victoria May 13 '15 at 12:49
• For that it's relatively simply Let us take your inequality $x-y > 3$ for example , and let's consider the points $(5,1)$ and $(3,1)$ If you plug in point one, you will get $5-1>3$ or $4>3$ , now is $4>3$? Of course it is hence the equality is true for $(5,1)$ , similarly for $3-1>3$ or $2>3$ , now is this true at all? NO! Hence the equality is false for $(3,1)$. In general you can do it for any point. @victoria – Mann May 13 '15 at 13:01

An equation (or, in your case, inequality) is not something that is true or false. An equation is something that can be either true or false, depending on the value of the variables that appear in the inequality.

For example, the inequality $\sin(x) > 0$ is true if $x=\frac\pi2$, but it is not true if $x=0$. Some inequalities, like $e^x<0$ are never true, and others, like $x^2\ge0$ are never false (for real values of $x$).

In your case, you probably need to find all of the values $(x,y)$ for which both your inequalities are true. To do that, I suggest you draw the sets \begin{align*} S_1 & = \{(x,y)| x+2y>6\}\\ S_2 & = \{(x,y)|x-y<3\} \end{align*}

to see where they intersect.

• I aligned the equations for sets $S_1$ and $S_2$ to make your answer more legible. You can see how I did it by right-clicking on the equations, then selecting Show Math As TeX Commands. – N. F. Taussig May 13 '15 at 15:23
• @N.F.Taussig Thanks for the edit. The tutorial is unnecesary, though (as 28k of experience should tell you :P) – 5xum May 13 '15 at 19:34

A mathematical inequality, just like any other mathematical statement using a relation, can be either true or false based on a set of criteria. If the statement contains only numbers and no variables, then it is either true or false, definitively. For example, we always know that $0<2$ is true, or $2<0$ is false.

However, say we define a statement with a variable with a relation, like for example $x=2$. If I were to ask you if that statement were true, you'd probably be puzzled. What do I mean, is this true? What I'm really asking is, for what values of $x$ is this statement true? Trivially, we can see that if $x$ is 2, then the statement is true and if $x$ is 3 or 4 or any other number, the statement is false.

We can extend this to your system of inequalities the exact same way. We can't ask "is this system true or false" because the answer to that depends on our choices of $x$ and $y$. There are some values of $x$ and $y$ that will make the given inequalities true, and there are some values of $x$ and $y$ that will make them false.

In this case, we can solve for $y$ and try to visualize the region of solutions. Here we find $y > x-3$ and $y > \dfrac{6-x}{2}$. So then any point that is greater than those lines will satisfy either equation. When you graph the equation and found the overlapping areas of inequality, you find where both inequalities were true, and thus the system. Notice that there are some areas where no points work, this is where the values of $x$ and $y$ will produce false mathematical statements. For example, $x = 3$, $y = -1$ makes the system

\begin{align*} -1 &> 3 \quad \text{False.} \\ -1 &> \frac{3}{2} \quad \text{False.} \end{align*}

It gets a little more complicated to find the areas with multiple variables and dimensions, but essentially the same principles hold. It is also useful to note that there are systems of equations (and inequalities) that are always true (eg. $y^2 \geq 0, x^2 \geq 0$), and systems that are always false. (eg. $y > 0, y \leq 0$.) But most of the ones you find in applications are false and true in different regions.

Hint: It would be helpful to sketch the lines $$l_1\colon \quad y=\frac{6-x}{2}\\ l_2\colon \quad y=x-3$$ and check when the following two conditions hold true jointly: $$y>\frac{6-x}{2}\\ y>x-3$$

• I just added a graph to my question if you would like to check it out. – victoria May 13 '15 at 12:53
• Well @victoria, I got confused before, yes, that's correct! – nullgeppetto May 13 '15 at 13:12