Graphing the Solution to Multiple Inequalities

In an earlier chapter we learned how to graph an inequality on a number line. Now we look at graphing inequalties on the coordinate plane.

To graph an inequalitiy we will use the same procedure as graphing an equation.

Simplify the inequality first then use the slope and intercept to graph.

2y > 4x - 6


2y > 4x - 6
2

y > 2x - 3

We anchor our line with the y-intercept and use the slope to extend the line in both directions.

With inequalities we use two different forms of a line:

dash or broken line for greater than or less than
or
solid line for great than or equal to or less than or equal to

But with inequalities we also shade in the section of the coordinate plane that includes all the possible values for y

If y is greater than 2x - 3 we shade in above the line.

For y is less than -x + 4 we shade below the line.

When we graph two inequalities we shade each individually and the area where they overlap is the solution to both inequalities

Important steps: simplify the inequality, graph the line using y-intercept and slope with either a dashed or solid line, shade above for greater and below for less.

Simultaneous Equations

Simultaneous equations - two or more equations that have one or more points (solutions) in common.


There are three ways to solve simultaneous equations: graphing, substitution and elimination.

Graphing:

To solve for a common solution to two equations using graphing, you graph each line either on paper or on a graphing calculator.

The point at which they intersect is the solution to both equations.

Example: Graph the solution for y = x + 1 and y = (-1/2)x + 4


The solution is the point at which the lines cross.


Some lines have more than one solution.

y = 2x - 2 and 3y = 6x - 6

Lines with the same slope and same y-intercept will be graphed one on top of the other so that all points coincide. All points on the lines are solutions (infinite solutions).



Lines with the same slope but different y-intercepts are parallel and have no points in common. In this situation there is no solution.


Substitution:

To use substitution you must have at least one equation in simplest form that is equal to one of the variables.

Example:
x + 2y = 8 and y = x + 1

Since the second equation is in simplest form (y =), we will use it to substitute.

If y = x + 1 we can substitute the value for y into the other equation:

x + 2(x + 1) = 8

Simplify using distribution:

x + 2x + 2 = 8

Combine like terms:

3x + 2 = 8

Simplify:

3x + 2 - 2 = 8 - 2

3x = 6

Divide both sides by 3.

x = 2

Now we need to solve for y too!

Use either of the original equations: x + 2y= 8 and y = x + 1

y = 2 + 1

y = 3

The solutions to this system of equations is ( 2, 3)




Elimination:

In this form we will eliminate one of the variable to solve for one at a time, similar to substitution but using addition and subtraction.

We will start with the same equations we used in the first two forms but written in Standard Form:

y = x + 1 becomes -x + y = 1

y = -1/2x +4 becomes 2y = -x = 8 when we clear the fraction,

then x + 2y = 8 in Standard Form.

Line the two equations up so that like terms are in a column:

-x + y = 1
x + 2y = 8

Combine like terms vertically.

-x and x cancel each other out eliminating x from the equation.

y + 2y = 3y

1 + 8 = 9

3y = 9

y = 3

Substitute the value for y back into either original equations:

-x + y = 1 or x + 2y = 8

x + 2(3) = 8

x + 6 = 8

x + 6 - 6 = 8 - 6

x = 2

The solution for these two equations is (2, 3).



For more complicated equations we may need to multiply one of the equations to make the coefficient inverse.

2x + 3y = 11
2x + y = 3

We can eliminate the x by multiplying the second equation by -1

2x + 3y = 11
-2x - y = - 3

The x is eliminated leaving:

3y = 11
- y = - 3

Combining those terms gives us

2y = 8

y = 4

Substituting y back in either of the original equations:

2x + 3(4) = 11

2x + 12 = 11

2x + 12 - 12 = 11 - 12

2x = -1

x = - 1/2

Solution (-1/2, 4)


Some equations are even more complicated and you must multiply to get one variable ready for elimination.

2x + 3y = 7
3x - y = 5

Multiply the second equation by 3 to eliminate the y.

2x + 3y = 7
9x -3y = 15

y is eliminated leaving

11x = 22

x = 2

Solve for y.

2(2) + 3y = 7

4 + 3y = 7

4 - 4 + 3y = 7 - 4

3y = 3

y = 1

Solution (2, 1)


For even more complicated equations you may need to multiply both equations.
5x + 2y = 1
2x - 3y = -11

We can multiply the first equation by 2 and the second equation by -5.

2(5x + 2y = 1)
-5(2x - 3y = -11)

10x + 4y = 2
-10x + 15y = 55

The x is eliminated leaving:

4y = 2
15y = 55

19y = 57

divide by 19

y = 3

Substitute the value for y back in either original equation.

5x + 2y = 1

5x + 2(3) = 1

5x + 6 = 1

5x + 6 - 6 = 1 - 6

5x = -5

x = -1

Solution (-1, 3)