Here's what this lesson offers:
You've already learned that the graph of $\,y = x + 1\,$
is the line shown at right.
This line is the picture of all the points $\,(x,y)\,$ that make the equation ‘$\,y=x+1\,$’ true. How can ‘$\,y=x+1\,$’ be true? For a given $\,x$value, the $\,y$value must equal $\,x + 1\,$. For each $\,x$value, there is exactly one corresponding $\,y$value—whatever $\,x\,$ is, plus $\,1\,$. The line is the picture of all the points $\,(x,x+1)\,$, as $\,x\,$ varies over all real numbers: $$ \cssId{s15}{\text{the line ‘$\,y=x+1\,$’ is all points of the form:}} \qquad (\ \ \cssId{s16}{x}\ \ , \cssId{s17}{\overbrace{x+1}^{\text{the $y$value EQUALS $x+1$}}}) $$ 
graph of $\,y = x + 1\,$: all points of the form $\,(x,x+1)$ 
Question:
What happens if the verb in the sentence ‘$\,y \color{red}{=} x+1\,$’
is changed from ‘$\,\color{red}{=}\,$’ to $\,\lt\,$, $\,\gt\,$, $\,\le\,$, or $\,\ge\,$?
Answer:
You go from a linear equation in two variables, to a linear inequality in two variables.
The solution set changes dramatically!
What was a line now becomes an entire halfplane:
graph of $\,y < x + 1\,$: all points of the form $\,(x,y)$ where the $y$value is less than $\,x+1\,$ line is dashed; shade below the line 
graph of $\,y > x + 1\,$: all points of the form $\,(x,y)$ where the $y$value is greater than $\,x+1\,$ line is dashed; shade above the line 
graph of $\,y \le x + 1\,$: all points of the form $\,(x,y)$ where the $y$value is less than or equal to $\,x+1\,$ line is solid; also shade below the line 
graph of $\,y \ge x + 1\,$: all points of the form $\,(x,y)$ where the $y$value is greater than or equal to $\,x+1\,$ line is solid; also shade above the line 
$3x  4y + 5 > 0$  $y \le 5x  1$ 
$x \ge 2\,$ (shorthand for $\,x + 0y \ge 2\,$) 
$y < 5\,$ (shorthand for $\,0x + y < 5\,$) 
So, what about graphing something like $\,2x  y < 3\,$, which isn't
in slopeintercept form?
You can (if desired) solve for $\,y\,$, and then use the method above: $$\begin{gather} \cssId{sb3}{2x  y < 3}\cr \cssId{sb4}{y < 2x + 3}\cr \cssId{sb5}{y > 2x  3} \end{gather} $$ (Remember to change the direction of the inequality symbol when you multiply/divide by a negative number.) The graph of $\,2x  y < 3\,$ is the same as the graph of $\,y > 2x  3\,$. Graph the line $\,y = 2x  3\,$ (dashed), and then shade everything above (see right). However, there's an easier way. Keep reading! 
graph of $\,2x  y < 3\,$ (which is equivalent to $\,y > 2x  3\,$) 
The ‘Test Point Method’ is socalled because it involves choosing a
‘test point’ to decide which side of the line to shade.
The process is illustrated with an example: graphing $\,2x  y < 3\,$.
The Test Point Method is usually easiest to use with sentences that aren't in slopeintercept form.
Step 1: IDENTIFICATION Recognize that ‘$\,2x  y < 3\,$’ is a linear inequality in two variables. Therefore, you know the graph is a halfplane. You need the boundary line; you need to know which side to shade. 

Step 2: BOUNDARY LINE Graph the boundary line $\,2x  y = 3\,$ using the intercept method. When $\,x = 0\,$, $\,y = 3\,$. When $\,y = 0\,$, $\,x = \frac{3}{2}\,$. Since the verb in ‘$\,2x  y < 3\,$’ is ‘$\,<\,$’, this line is not included in the solution set. Therefore, the line is dashed. 
graph the boundary line using the intercept method 
Step 3: TEST POINT TO DECIDE WHICH SIDE TO SHADE Choose a simple point that is NOT on the line. Whenever $\,(0,0)\,$ is available, choose it! Zeroes are very easy to work with. Is $\,(0,0)\,$ in the solution set? Substitute $\,x = 0\,$ and $\,y = 0\,$ into the original sentence ($\,2x  y < 3\,$), to see if it is true or false. Put a question mark over the inequality symbol, since you're asking a question: $$ \cssId{sb37}{2(0)  0 \overset{?}{<} 3} $$ If the result is TRUE, shade the side containing the test point. If the result is FALSE, shade the other side. Since ‘$\,0 < 3\,$’ is TRUE, shade the side containing $\,(0,0)\,$. Done! With so many zeroes involved in this method, computations can often be done in your head, making this QUICK and EASY! 
choose test point $\,(0,0)\,$: since ‘$2(0)  0 < 3\,$’ is TRUE, shade the side containing the test point 
Out of context, sentences like ‘$\,x \ge 2\,$’ and ‘$\,y < 5\,$’ can be confusing.
You only see one variable, but that doesn't necessarily mean that there isn't another variable with a zero coefficient!
Out of context, here are clues to what is likely wanted:
graph of $\,x\ge 2\,$, viewed as an inequality in ONE variable 
Viewed as an inequality in one variable, the solution set of ‘$\,x\ge 2\,$’ is the set of all numbers that are greater than or equal to $\,2\,$. The solution set is the interval $\,[2,\infty)\,$, shown at left. 
graph of $\,x\ge 2\,$, viewed as an inequality in TWO variables 
Viewed as an inequality in two variables, ‘$\,x\ge 2\,$’ is really a shorthand for ‘$\,x + 0y \ge 2\,$’. The solution set is the set of all points $\,(x,y)\,$, where the $x$value is greater than or equal to $\,2\,$. The $y$value can be anything! Here are examples of substitution into ‘$\,x + 0y \ge 2\,$’: The point $\,(2,5)\,$ is in the solution set, since ‘$\,2 + 0(5) \ge 2\,$’ is TRUE. The point $\,(3.5,7.4)\,$ is in the solution set, since ‘$\,3.5 + 0(7.4) \ge 2\,$’ is TRUE. The graph is the halfplane shown at left. This is the picture of all the points with $x$value greater than or equal to $\,2\,$. 
graph of $\,y\lt 5\,$, viewed as an inequality in TWO variables 
Viewed as an inequality in two variables, ‘$\,y\lt 5\,$’ is really a shorthand for ‘$\,0x + y \lt 5\,$’. The solution set is the set of all points $\,(x,y)\,$, where the $y$value is less than $\,5\,$. The $x$value can be anything! Here are examples of substitution into ‘$\,0x + y \lt 5\,$’: The point $\,(2,4)\,$ is in the solution set, since ‘$\,0(2) + 4 \lt 5\,$’ is TRUE. The point $\,(7.4,3)\,$ is in the solution set, since ‘$\,0(7.4) 3 \lt 5\,$’ is TRUE. The graph is the halfplane shown at left. This is the picture of all the points with $y$value less than $\,5\,$. 
CONCEPT QUESTIONS EXERCISE:
On this exercise, you will not key in your answer.However, you can check to see if your answer is correct. 
PROBLEM TYPES:
