Chapter+7

=​Chapter 7: Systems of Equations and Inequalities=

Chapter Overview
This chapter is about solving systems of linear equations and inequalities. This includes finding solutions to 2 or 3 equation systems, graphing inequalities, solving matrices and all the technical material in between. You will learn many methods to solve these systems of equations including graphing, substitution, elimination, determinents and matrices.

=Section 7.1: Two Equations Containing Two Variables =

__System of Equations:__
A system of equations is a collection of two or more equations, that each contain one or more variables.

__What is a solution of a system of equations:__
A solution to a system of equations is the value or values for the variables that satisfy all of the equations in the system. Systems of linear equations can have no solution (inconsistent), 1 solution (consistent and independent) or infinitely many solutions (consistent and dependent).

** __Methods of Solving Systems of Equations:__ **

 * 1) Graphing
 * 2) Substitution
 * 3) Elimination

This method is usually only used for finding the solution to simple systems of equations that are either already in slope-intercept form or can easily be changed into slope intercept form. Technically it can be used to solve any system of equation, but it is usually only used to find simple ones that are already in slope intercept form since it can get complicated. You use graphing by changing each equation in the system so that it is in slope intercept form. Then, you go ahead and graph the equations.If the lines of the equations intersect once, then it has one solution, if it is the same line, then it has infinitely many solutions and if the lines do no intersect at all, then they do not have any solutions and are inconsistent.
 * __Method #1: Graphing__ **

Here are some examples of graphs of systems of equations:



**__Example #1:__** Solving a system of linear equations by graphing

2x + y = 5 -4x + 6y = 12 Begin by changing both equations into slope-intercept form.

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

Now graph both lines in the xy coordinate plan.

The solution to this system is (3, -1).

The goal of this method is to create one equation with only one variable that is able to be solved. You do this by taking one of the equations in the system, solving for a variable, and plugging the solution into the other equation.
 * __Method #2: Substitution__ **

**__Example #2__**: Solve the system of linear equations by substitution.

**2**//**x**// **– 3**//**y**//**= –2 4**//**x**// **+** //**y**// ** = 24

Solve for one of the variables in one of the equations. It is often easiest to solve for the variable with no coefficient. **  4 // x // + // y // = 24 // y // = –4 // x // + 24 Now I'll plug this in ("substitute it") for " // y // " in the first equation, and solve for //x// : 2//x// – 3(–4//x// + 24) = –2 2//x// + 12//x// – 72 = –2 14//x// = 70 //x// = 5 Now plug the x value back into either equation to find the corresponding value of y. //y// = –4(5) + 24 = –20 + 24 = 4 **Then the solution is** **(**//**x**//**,** //**y**//**) = (5, 4)** **.**

Many students consider this method the easiest out of all of the methods simply because this method is the most universal. It can be used in any circumstance involving a system of equations. This method involves combining two of the equations in the system by either adding or subtracting them. By doing this you can eliminate at least one of the variables in the equation. If you do this enough times, then you will end up with an equation with only one variable that can easily be solved for then plugged back into the other equations to find the solution.
 * __Method #3: Elimination:__ **

Watch this great short video that demonstrates the elimination method.

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** __Example #3:__ ** solve this system of equations using elmination 2x-y=24 2x+2y=12

First figure out which variable you are going to eliminate and multiply or divide the equations so that the variable will be eliminated (2x-y=24)x2 4x-2y=48

Then, add the equations together to eliminate the variables 4x-2y=48 2x+2y=12 6x=60

Then, solve for the variable and plug it back into the equations to solve for the remaining variable 6x=60 x=10

2x-y=24 2(10)-y=24 20-y=24 y=20-24 y=-4

So your solutions are y=-4 and x=10 =Section 7.2: Systems of Linear Equations - 3 Equations Containing 3 Variables =

This section is much like section 7.1, the only difference is that in this section you are now solving systems of equations with three variables instead of two.

Possiblities for solutions: **__Example #1:__** Here is an example of how to solve one of these systems. Be sure to pay attention to the organization used to keep track of each equation and how they are being combined.

Dependent systems can get much more complicated. In dependent systems, you solve it like you would a normal three equations systems with three variables. The difference comes in later when you have your equations with two variables. From there, you solve for one of the variables for example, x. Then you solve for another one of the variables so that the two equations contain the same variable. For example, if you solved for x and the equation included y, then you would have to solve for z and your equation would once again contain y. Afterwards you write out the two equations as your answer and say that the last equation can be any number.
 * __Dependent Systems:__**

** __Example #2:__ ** Solve this system of equations: x-2y-z=8 Equation #1 2x-3y+z=23 Equation #2 4x-5y+5z=53 Equation #3

Combine equations 1 and 3 to eliminate z x-2y-z=8 2x-3y+z=23

3x-5y=31 Equation #4

Combine equations 1 and 2 to elminate z (x-2y-z=8)x5---> 5x-10y-5z=40 5x-10y-5z=40 4x-5y+5z=53

9x-15y=83 Equation #5

Combine equations 4 and 5 to eliminate y (3x-5y=31)x-3--> -9x+15y=-83 -9x+15y=-83 9x-15y=83 0=0 This means that it is a dependent system

take equation 4 and solve for x 3x-5y=31 3x=5y+31 x=5/3y+31/3<--- equation #6

take equation 1 and solve for z x-2y-z=8 z=x-2y-8 z=5/3y+31/3-2y-8<--- substitute equation #6 in for x z= -1/3y+7/3<-- equation #7

So your solutions are: x=5/3y+31/3 z=-1/3y+7/3 y= can be any number

= =

=Section 7.4: Systems of Linear Equations - Determinants =

Section 7.5: Matrix Algebra

When adding matrices the problems **are** communitive so A+B=B+A When subtracting matrices the problems **are not** communitive so A-B≠B-A Adding Examples: The numbers in the corresponding spot on the other matrix add together so 6+1=7, 2+5=7 , 4+4=8 , and 3+2=5
 * ADDING AND SUBTRACTING MATRICES **
 * *When adding and subtracting the matrices must have the same dimensions **

Subtraction Examples:

Scaler Multiplication: You distribute whatever number is in the front to the matrix
 * MULTIPLYING **

When multiplying matrices you have to make sure that the number of rows in one of the matrices is the same as the number of columns in the other ex: ** m × n n × r ** the n’s are the same so they can be multiplied and m and r will be the dimensions of the new matrix.

Multiplication Example: Solve A×B Since A is a 3x1 and B is a 1x3 they can be multiplied.

We can now solve for AxB. Since it is a 3x1 and a 1x3 it will be a 3x3. Steps to Solve 1. We multiply Column 1 of A by Row 1 of B for all values. 2x0=0, -5x0=0, 3x0=0 this is Column 1 of the new matrix 2. Now we multiply Column 2 of A by Row 2 of B for all values. 2x6=12, -5x6=-30, 3x6=18 this is Column 2 of the new matrix 3. Then we multiply Column 3 of A by Row 3 of B for all values. 2x-4=-8, -5x-4=20, 3x-4=-12 this is Column 3 of the new matrix The new matrix is… The identity matrices are the matrices that help you solve for the inverse of a matrix. The identity matrices are always square so the are nxn and have ones in the diagonal and zeros the rest. 2x2 Identity Matrix 3x3 Identity Matrix
 * IDENTITY MATRICES **

An ** identity matrix is equal to one ** because when you multiply a matrix by them you get one.

1. Take a matrix and multiply it by a variable matrix 2. Next use elimination to solve for the variables 3. Then plug in w and x to find y and z and those points are your inverse.
 * FINDING THE INVERSE OF A MATRIX: 2x2 **

Chapter 7.6 - Systems of Linear Inequalities; Linear Programming Graphing a Linear Inequality by hand- When graphing a linear equality, you first transform the equation to isolate y; 3x+ y £ 6, y £ 6-3x Now that you’ve isolated y, you can graph the equation- y=6-3x. ß Now that we’ve graphed the equation, we plug in a point to which side to shape of the function because the function is an inequality. A good place to start is always the **origin.** Also remember that when graphing inequalities, that if the inequality is strict ( < or > ), use a dashed line when graphing. If the equation is non-strict ( £ or ³ ), shown above. Graphing inequalities by calculator- First off, graph your equation as if the less than/greater than, /equal to sign is an equal sign ( = ). After you’ve completed that you pick a test point on the graph. For example- Graph the equation y=6-3x, than pick a point on the graph such as (-1,2). Than back out of the graph and type in the equation (2) £ 6-(-1) which equals- 2 £ 7; which is true; the calculator will show the answere as 1. If you pick another set of pints such as (5,5), the answere will will be zero because this equation doesn’t satisfy the function and the calculator’s answere will be 0.
 * (0,0)- 3(0) + (0) ****<span style="font-family: Symbol; mso-ascii-font-family: 'Times New Roman'; mso-char-type: symbol; mso-hansi-font-family: 'Times New Roman'; mso-symbol-font-family: Symbol; msoasciifontfamily: 'Times New Roman'; msochartype: symbol; msohansifontfamily: 'Times New Roman'; msosymbolfontfamily: Symbol;">£ ** 6, 0 <span style="font-family: Symbol; mso-ascii-font-family: 'Times New Roman'; mso-char-type: symbol; mso-hansi-font-family: 'Times New Roman'; mso-symbol-font-family: Symbol; msoasciifontfamily: 'Times New Roman'; msochartype: symbol; msohansifontfamily: 'Times New Roman'; msosymbolfontfamily: Symbol;">£ 6. This satisfies the function, therefore, we shade to the left of the line (where the origin is located). An easy way to think of it is, if it’s less-than, shade to the left; if it’s more than, shade to the right. *this graph is a half planes graph, because it’s divided into two regions. Here’s a link to a video on this subject- []

Graphing a system of linear inequalities- Graphing by hand- There will typically be 4 inequalities for each graph, in a system of inequalities. It’s easiest graph draw shade each one individually. For Example- x + y <span style="color: #333333; font-family: Symbol; font-size: 20pt; mso-ascii-font-family: 'Times New Roman'; mso-bidi-font-size: 12.0pt; mso-char-type: symbol; mso-hansi-font-family: 'Times New Roman'; mso-symbol-font-family: Symbol; msoasciifontfamily: 'Times New Roman'; msobidifontsize: 12.0pt; msochartype: symbol; msohansifontfamily: 'Times New Roman'; msosymbolfontfamily: Symbol;">£ 3, 2x + y <span style="color: #333333; font-family: Symbol; font-size: 20pt; mso-ascii-font-family: 'Times New Roman'; mso-bidi-font-size: 12.0pt; mso-char-type: symbol; mso-hansi-font-family: 'Times New Roman'; mso-symbol-font-family: Symbol; msoasciifontfamily: 'Times New Roman'; msobidifontsize: 12.0pt; msochartype: symbol; msohansifontfamily: 'Times New Roman'; msosymbolfontfamily: Symbol;">£ 4, x <span style="color: #333333; font-family: Symbol; font-size: 20pt; mso-ascii-font-family: 'Times New Roman'; mso-bidi-font-size: 12.0pt; mso-char-type: symbol; mso-hansi-font-family: 'Times New Roman'; mso-symbol-font-family: Symbol; msoasciifontfamily: 'Times New Roman'; msobidifontsize: 12.0pt; msochartype: symbol; msohansifontfamily: 'Times New Roman'; msosymbolfontfamily: Symbol;">³ 0, y <span style="color: #333333; font-family: Symbol; font-size: 20pt; mso-ascii-font-family: 'Times New Roman'; mso-bidi-font-size: 12.0pt; mso-char-type: symbol; mso-hansi-font-family: 'Times New Roman'; mso-symbol-font-family: Symbol; msoasciifontfamily: 'Times New Roman'; msobidifontsize: 12.0pt; msochartype: symbol; msohansifontfamily: 'Times New Roman'; msosymbolfontfamily: Symbol;">³  0. Sense both x & y are greater than 0, the graph must be in the **first quadrant**.

y <span style="font-family: Symbol; font-size: 20pt; mso-ascii-font-family: 'Times New Roman'; mso-bidi-font-size: 12.0pt; mso-char-type: symbol; mso-hansi-font-family: 'Times New Roman'; mso-symbol-font-family: Symbol; msoasciifontfamily: 'Times New Roman'; msobidifontsize: 12.0pt; msochartype: symbol; msohansifontfamily: 'Times New Roman'; msosymbolfontfamily: Symbol;">³ 0

x <span style="color: red; font-family: Symbol; font-size: 18pt; mso-ascii-font-family: 'Times New Roman'; mso-bidi-font-size: 12.0pt; mso-char-type: symbol; mso-hansi-font-family: 'Times New Roman'; mso-symbol-font-family: Symbol; msoasciifontfamily: 'Times New Roman'; msobidifontsize: 12.0pt; msochartype: symbol; msohansifontfamily: 'Times New Roman'; msosymbolfontfamily: Symbol;">³  0 y <span style="color: green; font-family: Symbol; font-size: 20pt; mso-ascii-font-family: 'Times New Roman'; mso-bidi-font-size: 12.0pt; mso-char-type: symbol; mso-hansi-font-family: 'Times New Roman'; mso-symbol-font-family: Symbol; msoasciifontfamily: 'Times New Roman'; msobidifontsize: 12.0pt; msochartype: symbol; msohansifontfamily: 'Times New Roman'; msosymbolfontfamily: Symbol;">£ 4-2x

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This relates to the next to segment of the chapter; which is minimizing and maximizing. Lastly, when minimizing and maximizing, you’re often given a system of four linear inequalities (shown above). The only difference between this and what I just did, is that you’re given an equation to maximize or minimize. Now, you must find the intersections- you can do this by using your graphing calculator; graph the functions and then press 2nd trace, 5. Than select a point above and below the intersection and then press enter. Or by simply setting the equations equal to each other and then using elimination as so- 4-2x=3-x; add x to each side and get- 4-x=3; then add another x and subtract 3 and you get - x=1. Now apply to the principle equation, which is z = 2x + 3y. The points of interest (intersections) are- (0,3) = 2(0) + 3(3) = 9 (1,2) = 2(1) + 3(2) = 8  (2,0) = 2(2) + 3(0) = 4

Nine is the maximum value of the equation and 4 is the minimum value of the equation. Here is the link to the correct chapter 7.6 wiki; I tried my best to put it on the website and the pictures didn't work, (as shown above).