# How do you graph the following?

## a) $\left\{\begin{matrix}y = x - 2 \\ y = - 2 x + 7\end{matrix}\right.$ b) $\left\{\begin{matrix}2 x + 4 y < 12 \\ x \ge - 2\end{matrix}\right.$

Sep 2, 2017

See explanation...

#### Explanation:

All of these equations and inequations in two variables are linear, having terms of no degree higher than $1$. So they correspond to straight lines. In example a) we just want to find the intersection of two lines. In example b) we want to identify the region bounded by two straight lines.

We can graph the lines by finding the intersections with the $x$ and $y$ axes.

Example a)

Given:

$y = x - 2$

Note that if we put $x = 0$ (or equivalently cover up the term involving $x$), then we get $y = - 2$. So the intersection with the $y$ axis is at $\left(0 , - 2\right)$.

If we put $y = 0$ then we get the equation $0 = x - 2$, from which we can deduce $x = 2$. So the intersection with the $x$ axis is at $\left(2 , 0\right)$

So we can draw the line $y = x - 2$ like this:
graph{(y-x+2)(x^2+(y+2)^2-0.01)((x-2)^2+y^2-0.01) = 0 [-10, 10, -5, 5]}

Given:

$y = - 2 x + 7$

Putting $x = 0$ we get $y = 7$. So the $y$ intercept is at $\left(0 , 7\right)$

Putting $y = 0$ we get $0 = - 2 x + 7$ and hence $x = \frac{7}{2}$. So the $x$ intercept is at $\left(\frac{7}{2} , 0\right)$.

Adding this line to our previous graph, it looks like this:
graph{(y-x+2)(x^2+(y+2)^2-0.01)((x-2)^2+y^2-0.01)(y+2x-7)(x^2+(y-7)^2-0.01)((x-7/2)^2+y^2-0.01)((x-3)^2+(y-1)^2-0.01) = 0 [-8.37, 11.63, -2.48, 7.52]}

The two lines appear to intersect at $\left(3 , 1\right)$, indicating the solution:

$x = 3 , y = 1$

If you want to check, substitute these values back into the original equations:

$y = 1 = 3 - 2 = x - 2$

$y = 1 = - 6 + 7 = - 2 \left(3\right) + 7 = - 2 x + 7$

$\textcolor{w h i t e}{}$
Example b)

Given:

$\left\{\begin{matrix}2 x + 4 y < 12 \\ x \ge - 2\end{matrix}\right.$

Using methods similar to example a), we can find the intercepts of the equation $2 x + 4 y = 12$ to be $\left(0 , 3\right)$ and $\left(6 , 0\right)$.

Then the region associated with the inequality $2 x + 4 y < 12$ lies entirely to one side of the line through these points. Which side? The side containing $\left(0 , 0\right)$, since $2 \left(2\right) + 4 \left(0\right) = 0 < 12$, so the origin satisfies the inequality.

graph{2x+4y < 12 [-8.37, 11.63, -2.48, 7.52]}

The second inequality $x \ge - 2$ tells us that we are only interested in the part of this region which lies on or to the right of the vertical line $x = - 2$. That is the intersection with:

graph{x>=-2 [-8.37, 11.63, -2.48, 7.52]}

So it looks something like this:

graph{sqrt(x+2)/sqrt(x+2)2x+4y < 12 [-8.37, 11.63, -2.48, 7.52]}