VOCABULARY, Simultaneous Equations and Inequalities(1)

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UNIT 5 VOCABULARY: SIMULTANEOUS EQUATIONS AND INEQUALITIES

1.1. Systems of linear equations

Sometimes you will be asked to find 2 unknown values by solving 2 equations at the same time. These types of equations are called simultaneous equations.

A system of linear equations with two unknows is an algebraic expression of this kind:

REMEMBER! LINEAR VS NON-LINEAR A Linear Equation has no exponent on any variable:

A solution to a system of linear equations is the ordered pair (x,y) that makes both equations true at the same time.

For example, for the system { x+y=32xy=0 , a solution would be the ordered pair (1, 2).

1.2. Solving by substitution and elimination

SUBSTITUTION

To explain this method, we are going to solve the following linear system: {3x+2y=19x+y=8These are the steps:

Write one of the equations so it is in the style "unknown = ..." You can start with any equation and any unknown. I will use the second equation and the variable "y" because it looks the simplest equation.

Now our system looks like: {3x+2y=19y=8x Replace (i.e. substitute) that variable in the other equation.

Now we replace "y" with "8 - x" in the other equation: 3x+2(8x)=19

Solve the other equation Using the usual algebra methods, we get 3x+162x=19 x=3


Find the value of the other variable

We know that x = 3, so if we replace "x" with "3" in the equation we first used:

y=8x y=83=5

ELIMINATION

Elimination is usually a faster method ... but needs to be kept neat. The idea is that you can safely do these:

You can multiply an equation by a constant (except zero), You can add (or subtract) an equation on to another equation

To explain this method, we are going to solve again the linear system: {3x+2y=19x+y=8Now ... my aim is to eliminate a variable from an equation.

First, I notice that there is a "2y" and a "y", so I have a clue how to proceed:

Multiply the second equation by two

{3x+2y=192x+2y=16 Subtract the second equation from the first equation:

x=3 Now we know what x is! Now, you can easily find the value of the other variable:

3+y=8 y=5

Exercises. Solve the following systems by both substitution and elimination:

a) {2x+3y=13x+4y=0 b) {x+y2 =x1xy2

=y+1c) { x+3y2 =53xy=5y d) { x+3y2 =542xy

2=1

2.1. Graphing method

The solution of a system of linear equations with two variables can be graphically represented as the point of intersection of two lines.

In order to do this, we use the cartesian coordinate system to refer to points in the 2-dimensional plane. For example, instead of writing x = 2 and y = 4, we can write (x, y) = (2, 4) or just refer to the point (2, 4).

To explain this method, we are going to solve again the linear system: {4xy=02x+y=6


y = 4x is simple to plot; it goes through the origin (0, 0), and when x = 1, y is y = 41 = 4, so the point is (1, 4).

2x + y = 6 is also simple. We can easily get that y = 6 2x, so when x = 0 the intercept is y = 6, and when y = 0 2x = 6 x = 3, so the line passes through the points (0,6) and (3,0).

2.2. Number of solutions

When the number of equations is the same as the number of variables, there is likely to be a solution, but not guaranteed.

In fact, there are three possible cases:

No solution One solution Infinite solutions

Here is a diagram for 2 equations in 2 variables:

Example 1.

{ 2xy=46x3y=3 Solving this system by elimination, multiplying by three the first equation, we get: {6x3y=126x3y=3 and in we subtract both equations 0 0 = 9 !!!!

What is going on here? Quite simply, there is no solution. They are actually parallel lines:


Example 2.

{ 2xy=46x3y=12 Solving this system by elimination, multiplying by three the first equation, we get: {6x3y=126x3y=12 and in we subtract both equations 0 0 = 0

What is going on here? Quite simply, there is no solution. They are actually parallel lines:

Well, that is actually TRUE! Zero does equal zero. That is because they are really the same equation (indeed, you can check that the second equation is the first equation but multiplied by three).

So there are an Infinite Number of Solutions.They are the same line:

Exercises. Solve graphically and by elimination, and classify:

a) { xy=2x+2y=11 b) {2x2y=0xy=2 c) { xy=22x2y=4 d) { 3x+y=4x4y=63.1. Non-linear systems

To solve a system of non-linear equations, we eliminate one of the unknowns, either by subtitution or elimination. Then, we substitute each value of the given equations and find the corresponding value(s) of the first unknow.

Example 1. {y+2x3=0y+x26=0


Step 1: In this example we eliminate y using the elimination method (the elimination method is the easiest one, but it is not always possible to use it).

We subtract both equations and we get: x2+2x+3=0

Step 2: we solve the resulting equation for the possible values of the unknow, which is a quadratic equation.

x2+2x+3=0 x=22241(3)21

=24

2 x1 = 3 and x2 = -1

Step 3: for each value found in step 2, we find the corresponding of the other unknown. For this, we substitute each value in one of the given equations (use the lowest degree equation if possible). In this case, we will use y + 2x 3 = 0.

x = 3 x = -1

y + 23 3 = 0 y + 2(-1) 3 = 0

y = -3 y = 5

Thus, the solutions to the given system of equations are (3, -3) and (-1, 5).

Example 2. {4x2+y2=13x+y=2 Step 1: Since we cannot eliminate x or y bu elimination, we must use the substitution method.

By solving the second equation for y, we get: y=2x . We substitute this expression for y in the first equation.

4x2+(2x)2=13 4x2+44x+x2=13 5x24x9=0

Step 2: we solve the resulting equation for the possible values of the unknow, which is a quadratic equation.

5x24x9=0 x=44245(9)25

=414

10 x1=

95

and x2 = -1

Step 3: for each value found in step 2, we find the corresponding of the other unknown. In this case, we will use x + y = 2.

x = 9/5 x = -1

9/5 + y = 2 -1 + y = 2

y = 1/5 y = 3

Thus, the solutions to the given system of equations are (9/5, 1/5) and (-1, 3).

Exercises. Solve the following systems of non-linear equations:

a) {x2+y2=25x+y=7 b) {x+y=7xy=12 c) {y22y+1=xx+y=5 d) { 1x2+ 1y2=131

x

1y=1


WORD PROBLEMS

1. What is the area of a rectangle knowing that its perimeter is 16 cm and its base is three times its height?

2. John bought a computer and a TV for 2,000 and later sold both items for 2,260. How much did each item cost, knowing that John sold the computer for 10% more than the purchase price, and the TV for 15% more?

3. A farm has pigs and turkeys, in total there are 58 heads and 168 paws. How many pigs and turkeys are there?

4. John says to Peter, "I have double the amount of money that you have" and Peter replies, "if you give me six dollars we will have the same amount of money". How much money does each have?

5. A company employs 60 people. Of this amount, 16% of the men wear glasses and 20% of the women also wear glasses. If the total number of people who wear glasses is 11, how many men and women are there in the company?

6. Find a pair of numbers using the follow conditions: If you add both numbers, the result will be 10. If you add one with two times the other, the result will be 17.

7. A group of students have paid 144 for 3 tickets in the shade and 6 tickets in the sun to a bull fight. Another group paid 66 for 2 tickets in the shade and 2 tickets in the sun. Calculate the price of each ticket.

8. Find the ages of two people knowing that ten years ago, the age of the first person was four times the age of the second person, and in twenty years, the age of the first person will be double the age of the second person.

9. A boy has a number of coins in both hands. If he passes two coins from the right to the left, there will be the same amount of coins in both hands. If he passes three coins from the left hand to the right hand, he will have double the amount of coins in his right hand than in his left hand. At the beginning, how many coins did the boy have in each hand?


4.1. Systems of inequalities: one variable

When solving systems of linear inequalities (or simultaneous inequalities) with one unknown, you must follow these steps:

Step 1: solve each inequality separately. Step 2: find the common values between the two inequalities (use a number line if

necessary).

For example, let us solve { x+22Step 1: We solve each inequality separately:

x + 2 < 6 2x 6 > - 2

x < -4 x > 2

So the common values are -4 < x < 2, which is the interval (-4, 2).

Exercises. Solve:

a) { 2x+31x+21 b) { 2x+31x+2


To do this, the easiest way is to choose a test point, and check if the inequality satisfy this point.

In this case, let's choose, for example, (0 , 0) as a test point. If we substitute the variables by the values, we get: -2x + y 3 -20 + 0 3 0 3

Is this true? Yes!!!

So the answer is the region that contains the point (0 , 0)

Another example. We are going to graph the solution to 2x 3y < 6First, I'll solve for y:

2x 3y = 6 3y = 6 2x 3y = 2x 6 y=2x6

3

We find a couple of points:

x 0 3

y -2 0But this exercise is what is called a "strict" inequality. That is, it isn't an "or equals to" inequality. In the case of these linear inequalities, the notation for a strict inequality is a dashed line. So the border of my solution region actually looks like this: By using a dashed line, I still know where the border is, but I also know that the border isn't included in the solution.

Using (0, 0) as a test point, I can easily find out that the final answer is:

A system of linear inequalities is a set of linear inequalities that you deal with all at once. Usually you start off with two or three linear inequalities. The technique for solving these systems is fairly simple. Here's an example.

Let's solve the following system: {2x3y12x+5y20x>0The first thing we have to do is to graph each individual inequality.


First inequality 2x 3y = 12 3y = 12 2x 3y = 2x 12

y=2x123

And using (0, 0) a a test point, we get the region above the line.

Second inequality

x + 5y = 20 5y = 20 x y=20x

5

And using (0, 0) a a test point, we get the region above the line.

Third inequality

The last inequality is a common "real life" constraint: only allowing x to be positive. The line "x = 0" is just the y-axis, and I want the right-hand side. I need to remember to dash the line in, because this isn't an "or equal to" inequality, the line isn't included in the solution:

The "solution" of the system is the region where all the inequalities are happy; that is, the solution is where all the inequalities work, the region where all three individual solution regions overlap. In this case, the solution is the shaded part in the middle:

Exercises. Solve:

a) {x4y2 b) { x+y02xy0 c) { x+y02xy0x6

Example 1.

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VOCABULARY, Simultaneous Equations and Inequalities(1)