# How do you graph the following?

##
**a)** #{ (y = x-2), (y = -2x+7) :}#

**b)** #{ (2x+4y < 12), (x >= -2) :}#

**a)**

**b)**

##### 1 Answer

#### Answer:

See explanation...

#### Explanation:

All of these equations and inequations in two variables are linear, having terms of no degree higher than

We can graph the lines by finding the intersections with the

**Example a)**

Given:

#y = x-2#

Note that if we put

If we put

So we can draw the line

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 = -2x+7#

Putting

Putting

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

#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(3)+7 = -2x+7#

**Example b)**

Given:

#{ (2x + 4y < 12), (x >= -2) :}#

Using methods similar to example a), we can find the intercepts of the *equation*

Then the region associated with the *inequality*

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

The second inequality

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]}