**Graphing Linear Equations and Inequalities:**

A solution for an equation with two variables is an ordered pair of real numbers that satisfy the equation. Consider the equation y = x + 3; for every number that is substituted into x a number y is produced and an ordered pair is formed.

Notice that there is an unlimited number of solutions to the equation.

This is the basis for plotting points or graphing an equation or inequality. Linear Equations have the basic form of Ax + By = C.

**Graphing Linear Equations:**

x and y-intercepts:

x-intercepts of a graph are the x-coordinates that are common with the x-axis.

To solve for the x-intercept, let y = 0, then solve for x.

y-intercepts of a graph are the y-coordinates that are common with the y-axis.

To solve for the y-intercept, let x = 0, then solve for y.

ex.

graph 2x -3y = 6

First, find the intercepts.

x = 0; -3y = 6

y=2; the point (0,-2) is on the line.y = 0; 2x = 6x = 3; the point (3,0) is on the line.

Now, check for a third point in the equation,

Let x = 6; 2(6) -3y = 612 -3y = 6-3y = -6y = 2

Graph y = 3x

x = 0; y = 3(0) = 0

The origin is part of the graph, and another two points must be found.

A table of values would be useful for this.

**Vertical and Horizontal Lines:**

x = 4, y = -5 are examples of vertical and horizontal equations of lines.

**Graphing Linear Inequalities:**

Graphing linear inequalities is almost the same as graphing linear equations, but with a slight difference.

ex.

graph 2x + 3y < 6

First graph the line 2x + 3y = 6, and use a broken line to graph the line since the line is not really part of the solution.

Choose a test point that is above and below the line to see if that point is a valid solution to the inequality. In this case, choose the origin (0,0) to see if it is a valid point in the solution.

2(0) + 3(0) < 60 < 6 the origin is part of thesolution.

Now, try a point above the line. Try point ( 5,4 ).

2(5) + 3(4) < 610 + 12 < 6 the point (5,4) is not part of the solution.

\ the solution graph is:

Any point in the blue area is a solution to the inequality.

Finding the Equation of a Line in the form Ax + B = C:

ex.

Find the equation of a line with a slope of -2 and a point at (4,6).

It is also known that a point (x,y) is on the line then,

From the formula:

Find the equation of the line that has the points (2,3) and (-5, 8)

First find the slope of the line.

Now that the slope is defined, the equation of the line can be found.

From the slope of the line, one of the given points and the variable point on the line (x,y); the equation of the line is from the formula;

Find the equation of the line with a slope of 3/2 and a y-intercept at 4.

The y-intercept implies that point (0,4) is in the line.

Now just solve as before:

3( x - 0 ) = 2( y - 4 )

3x = 2y - 8

3x - 2y = -8