SOLVING SYSTEMS OF INEQUALITIES

We solve systems of linear inequalities by GRAPHING.

Graph all of the inequalities on the same set of axes.
Find the area where the solutions overlap.

EX:

y ≥ 3x – 1
x < 4

First, graph each inequality on the same set of axes:

Here’s y ≥ 3x -1 without shading
(Remember the line is solid because it’s greater than or equal to)



Then we’ll graph x < 4 on the same set of axes, again without shading.

(Remember this line is dotted because x is less than but not equal to 4)



Now we need to look at the shading.

For y ≥ 3x – 1, let’s use test point (-1, 1)
Plug it into the inequality:

1≥ 3(-1) – 1
1≥ -3 – 1
1 ≥ -4

That’s true, so we shade on that side—the left side—of the line.


Let’s use that same test point for x < 4

-1 < 4

That’s also true, so we shade on the bottom side of that line.



In this system, we want to look for the places that BOTH inequalities are true—where the shading overlaps.

That’s in the green area below:



So our solution should look like this:





Note: You may have systems of inequalities with more than two inequalities. Then you want to find the area where all of them overlap.

GRAPHING LINEAR INEQUALITIES

SLOPE-INTERCEPT FORM

y = 2x -3

Here we have a slope of 2 and a y-intercept of -3. Let’s graph that line.


Now, let’s look at the inequality

y > 2x -3

Now y isn’t equal to 2x – 3, it’s greater than 2x – 3. To show that we don’t want to include the line y = 2x – 3, we turn our solid line into a dotted line.



Then, we shade in the part (on one side of the line) where y is greater than the line 2x – 3.

We can check if we should shade above or below by using test points. Pick a point on one side of the line, and a point on the other side.




Above, I picked (-1, 1) and (3, -1)

We test the points by plugging them into the inequality, y > 2x – 3.

(-1, 1): 1 > 2(-1) -3
1 > -2 -3
1 > -5 This is true, so we say that, yes,
at this point y is greater than 2x – 3.

(3, -1): -1 > 2(3) -3
-1 > 6 -3
-1 > 3 This is NOT true, so we say that,
no, at this point y is NOT greater than 2x -3.

So we shade in the side where (-1, 1) is:



And the inequality is graphed.

STANDARD FORM

3x – 2y ≥ 6

First, we graph the line 3x – 2y = 6 by finding the x- and y-intercepts.
x-int: 2, y-int: -3


In this case, because we have greater than OR equal to, we include the line. So the line stays solid.

Now we need to know which side to shade. We pick a point on one side of the line.

Let’s pick (0, 0). Plug it into the inequality, 3x – 2y ≥ 6

3(0) – 2(0) ≥ 6
0 – 0 ≥ 6
0 ≥ 6 This is false, so we don’t want to shade in
the side with (0, 0).

So we shade in the other side:

COMPOUND INEQUALITIES AND ABSOLUTE VALUES

A compound inequality is made up of two inequalities joined by “and” or “or”.

EX: And

x > -3 and x < 5

And means that both have to be true. That is, x must be both bigger than -3 and smaller than 5.

So if we were to graph the above inequality, it would look like this:




Notice, it’s the space between -3 and 5—where x is both bigger than -3 and less than 5.

EX: Or

x ≥ 4 or x < 0

Or means that at least one has to be true. That is, x must be either greater than 4 or less than 0.

So is we were to graph the above compound inequality, it would look like this:


So the solutions are everything except 0 and the numbers between 0 and 4.

You may also have to solve compound inequalities, like this:

EX: Solving

3x – 1 > 5 and –x – 6 > 1

You can solve each separately.
3x – 1 > 5
+1 +1
3x > 6
3 3

x > 2

-x + 6 > 1
-6 -6
-x > -5
x < 5

So we have x > 2 and x < 5


Sometimes, the inequalities are joined like this:

-3 < x + 5 ≤ 8

In these cases, we assume that we are using and instead of or.

In these cases, you can separate the two inequalities and solve them individually:

-3 < x + 5 and x + 5 ≤ 8

Or you can solve the two simultaneously—solve for x in the middle.

-3 < x + 5 ≤ 8
-5 -5 -5
-8 < x ≤ 3

ABSOLUTE VALUES

An absolute value |x| gives the positive value of x. For example:
|3| = 3 because 3 is already positive
|-3| = 3 as well.

When we have absolute values in inequalities, we end up with compound inequalities.

EX: Absolute value

|x| ≤ 4

Let’s think about this inequality. The x clearly can’t be anything bigger than 4. But it also can’t be anything smaller than -4, because the absolute value of something like -5 would be 5.

So we can rewrite this inequality as

x ≤ 4 and x ≥ -4
or you can write it as -4 ≤ x ≤ 4

EX: Solving with an absolute value

5 < |x| - 2
+2 +2
7 < |x| Now we use that trick
-7 < x < 7

EX: Solving with an absolute value

4 < |x – 1| Here, we use the absolute value trick first, since the 1 is inside the absolute value
-4 < x – 1 < 4 Then solve.

Solving Systems by Graphing

We will use the following system.

y	=	-2 x	- 4
y	=	1 x	+ 5
		4

Once you know how to graph, solving by graphing is easy.

1—Graph the two lines.

2—See where the two lines intersect.

Here, they intersect at (-4, 4)

That point is the SOLUTION.

3—Check your answer by plugging it in to the equations.

(You have to plug it into BOTH equations, since you are checking whether it is the solution to the SYSTEM, not to just one or the other of the equations.)

4	=	-2 (-4)	- 4
4	=	8	- 4	That’s true, so (-4, 4) is a solution of the first equation
4	=	1 (-4)	+ 5
		4
4	=	-1	+5	That’s true, so (-4, 4) is a solution of the second equation.

Since (-4, 4) is a solution to both equations, it is the SOLUTION of the SYSTEM.

And that is how you solve a system by graphing.

Solving Systems by Elimination

OK, OK. Now this is the third way we can solve systems of equations. It’s exciting!!!

Remember how you can do ANYTHING you want (except divide by 0) to one side of an equation, as long as you do the same thing to the other side of the equation?

Well, we can use this trick to add two equations together.

Take, for example, the following system of equations:

3x – 4y = -2

2x + 4y = -3

If we take 3x – 4y = -2, we can add 2x + 4y to one side of the equation, and 3 to the other side because 2x + 4y and 3 are equal. So we’re actually adding the same thing to both sides!

We add straight down, combining like terms.

3x	-4y	=	-2
+2x	+4y		+(-3)
5x	+0	=	-5
5x		=	-5	Woah, check it out! One of our variables is gone! Now we can solve for x!
x		=	-1

Cool. Now to find y, we just substititute -1 for x in one of the equations.

3(-1)	-4y	=	-2	Simplify
-3	-4y	=	-2
+3			+3	Subtract 3 to get the y’s alone
	-4y	=	1	And divide by -4 to get the y by itself
	-4		-4
	y	=	-1 4	Hooray! We’re done!

So our solution is (-1, -1/4)

We can either add or subtract our two equations, depending upon what will cancel out one of the variables.

MULTIPLYING TO ELIMINATE

Now, let’s look at an example where it’s a little bit harder to solve by elimination.

3x + 5y = 7

4x + 10y = 16

Oh great, there are no coefficients that are the same! How will we ever cancel things out!?!

Luckily, we can use that same super-useful math fact, that we can do whatever we want to an equation as long as we do it to both sides.

Watch what happens if we multiply the first equation by -2…

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

-6x – 10y = -14

Now the coefficients of y will cancel! We replace the first equation with the multiplied by -2 equation we just made and solve by elimination!

-6x	-10y	=	-14	Add down
+4x	+10y	=	+16	Combine like terms
-2x	+0	=	2
-2x		=	2	Hooray! Only one variable! Let’s solve!
x		=	-1

Now we substitute that x = -1 back into one of our original equations, and we get:

3(-1)	+ 5y	=	7	Simplify
-3	+ 5y	=	7
+3			+3	Solve for y by getting the y’s alone
	5y	=	10	Then dividing by 5 to get y by itself
	5		5
	y	=	2	Hooray! Now we know x and y!

So our solution is (-1, 2)

Systems of Equations and Solutions

SYSTEMS OF EQUATIONS AND SOLUTIONS

SYSTEM OF EQUATIONS

A “SYSTEM” is a set of equations with the same variables.

EX:

x - 3y = 12

2x + 4y = 4

We’re looking at systems of linear equations—equations of lines.

SOLUTIONS

The SOLUTION of a system of equations is made up of solutions that work for both equations.

EX: The solution to the equations above is x = 6, y = -2) because that point fits into both equations:

Plugging in 6 for x and -2 for y in the first equation, we get

6 – 3(-2) = 12

6 + 6 = 12

12 = 12

Plugging in 6 for x and -2 for y in the second equation, we get

2(6) + 4(-2) = 4

12 – 8 = 4

4 = 4

GRAPHING SYSTEMS

We can visualize systems by graphing the equations on the same set of axes. The solution is where the lines cross.

This is the graph of the example above. You can see the lines cross at (6, -2).

A system of equations may have ONE solution, MANY solutions, or NO SOLUTIONS.

If the lines cross ONCE, there is ONE specific solution.

If the lines are the SAME, they are crossing many, many times (at every single point on the line), so there are infinitely MANY SOLUTIONS.

If the lines are PARALLEL, there are NO SOLUTIONS, because the lines never cross.

(Note that these lines have the same slope but different intercepts.)

Graphing from Standard Form

STANDARD FORM

Standard form of the equation of a line looks like this:

Ax + By = C
We want A, B, and C to be whole numbers, and we want A to be positive.

We can graph from standard form by using the x- and y-intercepts.


INTERCEPTS

The x-intercept is where the line crosses the x-axis.
The y-intercept is where the line crosses the y-axis.

Note that at the x-intercept, y is zero. (We haven't moved up or down from the x-axis.)
At the y-intercept, x is zero. (We haven't moved right or left from the y-axis.)


GRAPHING FROM STANDARD FORM

Ex: 3x-4y=12

To find the x-intercept, we plug in 0 for y.

3x-4(0)-12 ..........................4(0) becomes 0
3x=12 ................................ now divide both sides by 3
x=4

So 4 is the x-intercept.

To find the y-intercept, we plug in zero for x.

3(0)-4x-12 ..........................3(0) becomes 0
-4x=12 ................................now divide both sides by -4
x=-3

So -3 is the y-intercept.

Graph the two intercepts and draw a line through them.