Graph Solution On Number Line

Systems of Equations and Inequalities

In previous capacity we solved equations with one unknown or variable. Nosotros will now study methods of solving systems of equations consisting of two equations and two variables.

POINTS ON THE PLANE

OBJECTIVES

Upon completing this section y'all should be able to:

Correspond the Cartesian coordinate system and identify the origin and axes.
Given an ordered pair, locate that signal on the Cartesian coordinate organization.
Given a bespeak on the Cartesian coordinate system, state the ordered pair associated with it.

We take already used the number line on which we have represented numbers every bit points on a line.

Note that this concept contains elements from two fields of mathematics, the line from geometry and the numbers from algebra. Rene Descartes (1596-1650) devised a method of relating points on a plane to algebraic numbers. This scheme is chosen the Cartesian coordinate system (for Descartes) and is sometimes referred to as the rectangular coordinate system.

This system is equanimous of two number lines that are perpendicular at their zero points.

Perpendicular means that two lines are at right angles to each other.

Report the diagram advisedly equally you note each of the following facts.

The number lines are called axes. The horizontal line is the 10-axis and the vertical is the y-axis. The zero bespeak at which they are perpendicular is called the origin.

Axes is plural. Axis is atypical.

Positive is to the right and upwards; negative is to the left and down.

The arrows indicate the number lines extend indefinitely. Thus the plane extends indefinitely in all directions.

The plane is divided into four parts chosen quadrants. These are numbered in a counterclockwise direction starting at the upper correct.

Points on the airplane are designated by ordered pairs of numbers written in parentheses with a comma between them, such as (5,7). This is called an ordered pair because the social club in which the numbers are written is of import. The ordered pair (5,7) is not the same as the ordered pair (seven,v). Points are located on the plane in the following mode.

Offset, offset at the origin and count left or right the number of spaces designated past the first number of the ordered pair. Second, from the betoken on the x-axis given past the first number count upward or down the number of spaces designated by the second number of the ordered pair. Ordered pairs are always written with x first and and so y, (ten,y). The numbers represented by x and y are called the coordinates of the point (x,y).

This is important. The outset number of the ordered pair always refers to the horizontal management and the 2nd number always refers to the vertical direction.

Instance one On the following Cartesian coordinate system the points A (iii,four), B (0,5), C (-2,7), D (-iv,i), E (-three,-4), F (iv,-two), 1000 (0,-v), and H (-half-dozen,0) are designated. Check each one to determine how they are located.

What are the coordinates of the origin?

GRAPHING LINEAR EQUATIONS

OBJECTIVES

Upon completing this section you should be able to:

Find several ordered pairs that make a given linear equation truthful.
Locate these points on the Cartesian coordinate organisation.
Depict a straight line through those points that stand for the graph of this equation.

A graph is a pictorial representation of numbered facts. In that location are many types of graphs, such equally bar graphs, circular graphs, line graphs, and then on. Y'all can commonly find examples of these graphs in the financial section of a newspaper. Graphs are used considering a picture usually makes the number facts more hands understood.

In this department we will discuss the method of graphing an equation in two variables. In other words, nosotros volition sketch a motion-picture show of an equation in two variables.
Consider the equation x + y - 7 and note that nosotros can easily observe many solutions. For instance, if ten = 5 and then y - two, since five + 2 = 7. Also, if x = 3 then y = four, since 3 + 4 = seven. If nosotros represent these answers as ordered pairs (x,y), so we have (5,2) and (three,4) as two points on the aeroplane that correspond answers to the equation ten + y = seven.

All possible answers to this equation, located as points on the airplane, will requite the states the graph (or picture) of the equation.

Of class we could never find all numbers x and y such that x + y = 7, so we must be content with a sketch of the graph. A sketch can be described as the "bend of best fit." In other words, it is necessary to locate enough points to give a reasonably accurate flick of the equation.

Retrieve, in that location are infinitely many ordered pairs that would satisfy the equation.

Example one Sketch the graph of 2x + y = 3.

Solution We wish to notice several pairs of numbers that volition make this equation true. We will attain this by choosing a number for x and so finding a corresponding value for y. A table of values is used to tape the data.

In the pinnacle line (x) nosotros will place numbers that nosotros have chosen for x. Then in the lesser line (y) we volition place the corresponding value of y derived from the equation.

Of course, we could as well start by choosing values for y and then find the corresponding values for x.

In this instance nosotros will allow x to have on the values -three, -2, -ane,0, 1,two,three.

These values are capricious. We could choose any values at all.

Notice that once we have chosen a value for x, the value for y is determined by using the equation.

These values of x give integers for values of y. Thus they are practiced choices. Suppose we chose

These facts give us the post-obit tabular array of values:

We now locate the ordered pairs (-iii,ix), (-2,7), (-one,5), (0,3), (1,1), (2,-1), (3,-3) on the coordinate plane and connect them with a line.

We now have the graph of 2x + y = three.

The line indicates that all points on the line satisfy the equation, as well as the points from the table. The arrows indicate the line continues indefinitely.

The graphs of all first-caste equations in two variables will exist straight lines. This fact will exist used here fifty-fifty though it will be much afterward in mathematics earlier you tin can prove this argument. Such first-degree equations are called linear equations.

Thus, any equation of the form ax + by - c where a, b, and c are real numbers is a linear equation.

Equations in two unknowns that are of college degree give graphs that are curves of different kinds. Y'all will study these in future algebra courses.

Since the graph of a beginning-degree equation in two variables is a direct line, it is only necessary to have two points. Still, your work volition exist more consistently accurate if you notice at least iii points. Mistakes can be located and corrected when the points found do not prevarication on a line. Nosotros thus refer to the third point as a "checkpoint."

This is important. Don't attempt to shorten your work by finding only two points. You will be surprised how often yous will find an mistake by locating all three points.

Case 2 Sketch the graph of 3x - 2y - 7.

Solution First make a tabular array of values and decide on three numbers to substitute for ten. We volition try 0, ane,2.

Again, you could also accept started with capricious values of y.

The answer is not every bit like shooting fish in a barrel to locate on the graph equally an integer would be. So information technology seems that x = 0 was not a very adept selection. Sometimes it is possible to look alee and make better choices for 10.

Since both x and y are integers, x = 1 was a expert selection.

The point (1,-2) will be easier to locate. If x = 2, we will take another fraction.

The signal (3,1) will exist easy to locate.

x = 3 was some other practiced choice.

Nosotros will readjust the table of values and apply the points that gave integers. This may not always be feasible, but trying for integral values will give a more accurate sketch. Nosotros now have the table for 3x - 2y = seven.

Nosotros can exercise this since the choices for x were arbitrary.

Locating the points (1,-two), (3,1), (- 1,-5) gives the graph of 3x - 2y = 7.

How many ordered pairs satisfy this equation?

SLOPE OF A LINE

OBJECTIVES

Upon completing this section you should be able to:

Associate the slope of a line with its steepness.
Write the equation of a line in slope-intercept course.
Graph a directly line using its gradient and y-intercept.

Nosotros now wish to talk over an of import concept chosen the slope of a line. Intuitively we can think of slope as the steepness of the line in relationship to the horizontal.

Post-obit are graphs of several lines. Written report them closely and mentally reply the questions that follow.

Which line is steeper?

What seems to be the relationship between the coefficient of x and the steepness Which graph would be steeper: of the line when the equation is of the class y = mx?

Which graph would be steeper: y = 3x or y = 7x?

Now report the following graphs.

Which line is steeper?

What effect does a negative value for m have on the graph?

Which graph would be steeper: y = 3x or y = 7x?

For the graph of y = mx, the following observations should have been made.

If k > 0, then
- equally the value of m increases, the steepness of the line increases and
- the line rises to the correct and falls to the left.
If thousand < 0, then
- as the value of 1000 increases, the steepness of the line decreases and
- the line rises to the left and falls to the correct

Call back, thousand > 0 ways "one thousand is greater than zero."

In other words, in an equation of the form y - mx, m controls the steepness of the line. In mathematics we utilise the word slope in referring to steepness and course the post-obit definition:

In an equation of the course y = mx, m is the slope of the graph of the equation.

Case 1 Sketch the graph of y = 6x and give the gradient of the line.

Solution Nosotros beginning make a tabular array showing three sets of ordered pairs that satisfy the equation.

Remember, nosotros only need 2 points to determine the line only we employ the third point as a check.

We then sketch the graph.

The value of m is six, therefore the slope is 6. We may merely write thousand - six.

Case ii Sketch the graph and state the slope of

Solution Choosing values of ten that are divisible by 3, nosotros obtain the table

Why use values that are divisible by 3?

Then the graph is

The gradient of

We at present wish to compare the graphs of two equations to establish another concept.

Instance 3 Sketch the graphs of y 3x and y - 3x + 2 on the same set of coordinate axes.

Compare the coefficients of x in these two equations.

Solution

In example 3 look at the tables of values and annotation that for a given value of x, the value of y in the equation y = 3x + ii is two more than the corresponding value of y in the equation y = 3x.

Look now at the graphs of the two equations and note that the graph of y = 3x + 2 seems to have the same gradient as y = 3x. Also notation that if the unabridged graph of y = 3x is moved upwardly ii units, information technology will be identical with the graph of y = 3x + 2. The graph of y = 3x crosses the y-axis at the point (0,0), while the graph of y = 3x + 2 crosses the y-centrality at the indicate (0,2).

Over again, compare the coefficients of ten in the two equations.

Compare these tables and graphs equally in example iii.

Observe that when two lines have the same slope, they are parallel.

The gradient from one indicate on a line to another is determined by the ratio of the modify in y to the modify in ten. That is,

Notation that the modify in x is 3 and the modify in y is two.

The change in x is -4 and the change in y is 1.

Nosotros could besides say that the modify in x is iv and the change in y is - one. This volition consequence in the aforementioned line.

Example seven In the graph of y = 3x - 2 the slope is 3.

The change in x is 1 and the alter in y is iii.

y = mx + b is called the gradient-intercept form of the equation of a straight line. If an equation is in this form, m is the gradient of the line and (0,b) is the signal at which the graph intercepts (crosses) the y-axis.

The point (0,b) is referred to as the y-intercept.

If the equation of a straight line is in the slope-intercept class, it is possible to sketch its graph without making a table of values. Use the y-intercept and the slope to draw the graph, equally shown in case 8.

Annotation that this equation is in the class y = mx + b.

First locate the point (0,-2). This is one of the points on the line. The slope indicates that the changes in x is 4, so from the betoken (0,-2) we move four units in the positive management parallel to the x-axis. Since the change in y is 3, nosotros and then move iii units in the positive management parallel to the y-axis. The resulting betoken is also on the line. Since two points make up one's mind a straight line, we so depict the graph.

Always kickoff from the y-intercept.
A common error that many students make is to confuse the y-intercept with the x-intercept (the signal where the line crosses the x-axis).

Example 9 Give the slope and y-intercept and sketch the graph of y = 3x + 4.

Solution chiliad = -3, y-intercept = (0,iv).

To express the slope as a ratio we may write -three as or . If we write the slope as , then from the signal (0,iv) we movement one unit in the positive direction parallel to the x-axis and then move iii units in the negative management parallel to the y-axis. Then we describe a line through this point and (0,4).

Suppose an equation is non in the form y = mx + b. Can we nonetheless find the gradient and y-intercept? The reply to this question is yes. To practice this, nevertheless, we must alter the form of the given equation by applying the methods used in section 4-2.

Department 4-2 dealt with solving literal equations. You may want to review that department.

Example 10 Find the slope and y-intercept of 3x + 4y = 12.

Solution First nosotros recognize that the equation is not in the gradient-intercept form needed to answer the questions asked. To obtain this form solve the given equation for y.

Example 11 Find the slope and y-intercept of 2x - y = 7.

Solution Placing the equation in slope-intercept grade, we obtain

Sketch the graph of the line on the grid below.

GRAPHING LINEAR INEQUALITIES

OBJECTIVES

Upon completing this section you should exist able to graph linear inequalities.

In chapter 4 we synthetic line graphs of inequalities such every bit

These were inequalities involving only one variable. We found that in all such cases the graph was some portion of the number line. Since an equation in two variables gives a graph on the plane, it seems reasonable to assume that an inequality in two variables would graph as some portion or region of the plane. This is in fact the instance. The solution of the inequality x + y < 5 is the set of all ordered pairs of numbers {x,y) such that their sum is less than v. (x + y < 5 is a linear inequality since x + y = 5 is a linear equation.)

Example one Are each of the following pairs of numbers in the solution set of x + y < 5? (2,one), (3,-4), (v,vi), (3,2), (0,0), (-1,4), (-ii,8).

Solution

The solution set consists of all ordered pairs that make the statement truthful.

To summarize, the post-obit ordered pairs give a true statement.
(2,i),(3,-four),(0,0),(-ane,4)

The following ordered pairs give a imitation statement.
(5,half-dozen),(3,2),(-2,8)

Following is a graph of the line x + y = 5. The points from example 1 are indicated on the graph with answers to the question "Is x + y < v?"

Notice that all the points that satisfy the equation are to the left and below the line while all the points that will not are in a higher place and to the right.

Observe that all "yes" answers lie on the same side of the line x + y = 5, and all "no" answers lie on the other side of the line or on the line itself.

The graph of the line x + y = 5 divides the plane into 3 parts: the line itself and the ii sides of the lines (called one-half-planes).

ten + y < 5 is a one-half-plane
x + y < 5 is a line and a half-plane.

If ane signal of a one-half-airplane is in the solution set of a linear inequality, then all points in that half-plane are in the solution prepare. This gives u.s. a convenient method for graphing linear inequalities.

To graph a linear inequality
ane. Replace the inequality symbol with an equal sign and graph the resulting line.
2. Check one signal that is obviously in a particular half-airplane of that line to come across if it is in the solution set of the inequality.
iii. If the point chosen is in the solution set, then that unabridged half-aeroplane is the solution gear up. If the point chosen is non in the solution set, then the other half-airplane is the solution prepare.

Why do we need to check only one bespeak?

Example 2 Sketch the graph of 2x 4- 3y > 7.

Solution Step one: Outset sketch the graph of the line 2x + 3y = 7 using a table of values or the slope-intercept form.

Step ii: Next cull a point that is not on the line 2x + 3y = 7. [If the line does not go through the origin, then the point (0,0) is e'er a skillful selection.] Now turn to the inequality 2x + 3y> > 7 to meet if the chosen point is in the solution set.

Step 3: The bespeak (0,0) is not in the solution set, therefore the one-half-plane containing (0,0) is not the solution fix. Hence, the other halfplane determined by the line 2x + 3y = vii is the solution set.
Since the line itself is not a part of the solution, information technology is shown as a dashed line and the one-half-plane is shaded to show the solution fix.

The solution set up is the one-half-plane above and to the right of the line.

Example 3 Graph the solution for the linear inequality 2x - y ≥ four.

Solution Stride ane: Kickoff graph 2x - y = 4. Since the line graph for 2x - y = 4 does not go through the origin (0,0), cheque that bespeak in the linear inequality.

Step 2:

Step 3: Since the point (0,0) is non in the solution set, the half-plane containing (0,0) is not in the gear up. Hence, the solution is the other one-half-plane. Notice, yet, that the line 2x - y = 4 is included in the solution set. Therefore, draw a solid line to show that it is part of the graph.

The solution set is the line and the one-half-plane below and to the right of the line.

Example 4 Graph ten < y.

Solution Outset graph ten = y. Side by side check a point not on the line. Notice that the graph of the line contains the indicate (0,0), so we cannot utilize it equally a checkpoint. To determine which half-plane is the solution fix use any point that is plain not on the line 10 = y. The indicate ( - 2,3) is such a indicate.

Using this information, graph ten < y.

When the graph of the line goes through the origin, any other point on the ten- or y-axis would also be a good choice.

GRAPHICAL SOLUTION OF A System OF LINEAR EQUATIONS

OBJECTIVES

Upon completing this section you should be able to:

Sketch the graphs of two linear equations on the same coordinate organization.
Determine the mutual solution of the two graphs.

Instance ane The pair of equations is called a system of linear equations.

Nosotros take observed that each of these equations has infinitely many solutions and each will form a straight line when we graph it on the Cartesian coordinate system.

We now wish to find solutions to the arrangement. In other words, we want all points (x,y) that will exist on the graph of both equations.

Solution We reason in this manner: If all solutions of 2x - y = 2 lie on i straight line and all solutions of 10 + 2y = 11 lie on another direct line, then a solution to both equations will be their points of intersection (if the 2 lines intersect).

In this table we allow x take on the values 0, 1, and two. Nosotros and so find the values for y by using the equation. Exercise this before going on.
In this tabular array we let y take on the values 2, iii, and 6. We then find x by using the equation. Check these values also.

The two lines intersect at the point (iii,four).

Note that the bespeak of intersection appears to be (3,4). We must now check the point (3,4) in both equations to encounter that it is a solution to the organization.

As a bank check we substitute the ordered pair (3,4) in each equation to run into if we get a true statement.
Are there whatsoever other points that would satisfy both equations? Why?

Therefore, (iii,4) is a solution to the system.

Not all pairs of equations volition requite a unique solution, as in this example. In that location are, in fact, three possibilities and you should be aware of them.

Since we are dealing with equations that graph as straight lines, we can examine these possibilities by observing graphs.

1. Contained equations The 2 lines intersect in a single betoken. In this example there is a unique solution.

The instance above was a organisation of independent equations.

2. Inconsistent equations The 2 lines are parallel. In this case there is no solution.

No thing how far these lines are extended, they will never intersect.

3. Dependent equations The ii equations give the same line. In this case whatever solution of one equation is a solution of the other.

In this case in that location volition be infinitely many common solutions.

In later algebra courses, methods of recognizing inconsistent and dependent equations will be learned. However, at this level we will bargain only with independent equations. You lot can then expect that all problems given in this chapter will take unique solutions.

This means the graphs of all systems in this chapter will intersect in a single indicate.

To solve a system of two linear equations by graphing
one. Make a table of values and sketch the graph of each equation on the same coordinate arrangement.
2. Notice the values of (10,y) that name the point of intersection of the lines.
iii. Check this indicate (x,y) in both equations.

Again, in this table wc arbitrarily selected the values of x to be - 2, 0, and 5.
Here we selected values for x to be two, 4, and vi. You lot could have chosen whatsoever values you wanted.
We say "apparent" considering nosotros have not yet checked the ordered pair in both equations. Once it checks it is so definitely the solution.

Since (3,2) checks in both equations, it is the solution to the organization.

GRAPHICAL SOLUTION OF A Organization OF LINEAR INEQUALITIES

OBJECTIVES

Upon completing this section y'all should be able to:

Graph ii or more linear inequalities on the same set of coordinate axes.
Decide the region of the plane that is the solution of the arrangement.

Subsequently studies in mathematics will include the topic of linear programming. Even though the topic itself is beyond the scope of this text, one technique used in linear programming is well inside your reach-the graphing of systems of linear inequalities-and we will discuss information technology here.

You lot found in the previous department that the solution to a organization of linear equations is the intersection of the solutions to each of the equations. In the aforementioned fashion the solution to a arrangement of linear inequalities is the intersection of the one-half-planes (and perhaps lines) that are solutions to each individual linear inequality.

In other words, x + y > 5 has a solution set and 2x - y < four has a solution set. Therefore, the organization

has as its solution set the region of the aeroplane that is in the solution gear up of both inequalities.

To graph the solution to this system nosotros graph each linear inequality on the same set of coordinate axes and bespeak the intersection of the two solution sets.

Note that the solution to a arrangement of linear inequalities will be a collection of points.

Over again, utilise either a table of values or the slope-intercept form of the equation to graph the lines.

Checking the indicate (0,0) in the inequality 10 + y > 5 indicates that the indicate (0,0) is not in its solution set. Nosotros indicate the solution set up of x + y > 5 with a screen to the correct of the dashed line.

This region is to the correct and above the line x + y = 5.

Checking the indicate (0,0) in the inequality 2x - y < iv indicates that the indicate (0,0) is in its solution set. We indicate this solution prepare with a screen to the left of the dashed line.

This region is to the left and higher up the line 2x - y = 4.

The intersection of the two solution sets is that region of the airplane in which the two screens intersect. This region is shown in the graph.

Notation again that the solution does non include the lines. If, for example, we were asked to graph the solution of the arrangement

which indicates the solution includes points on the line x+ y = five.

The results indicate that all points in the shaded section of the graph would be in the solution sets of 10 + y > 5 and 2x - y < four at the same time.

SOLVING A Arrangement Past SUBSTITUTION

OBJECTIVES

Upon completing this section you lot should be able to solve a organisation of ii linear equations by the substitution method.

In department half-dozen-5 we solved a arrangement of two equations with two unknowns by graphing. The graphical method is very useful, but it would not exist practical if the solutions were fractions. The bodily point of intersection could exist very difficult to make up one's mind.
At that place are algebraic methods of solving systems. In this section nosotros will discuss the method of substitution.

Case i Solve by the commutation method:

Solution
Step one We must solve for one unknown in one equation. Nosotros tin choose either ten or y in either the first or second equation. Our pick can be based on obtaining the simplest expression. In this example nosotros will solve for x in the 2d equation, obtaining x = 4 + 2y, considering any other choice would accept resulted in a fraction.

Look at both equations and see if either of them has a variable with a coefficient of one.

Step 2 Substitute the value of x into the other equation. In this case the equation is
2x + 3y = 1.
Substituting (4 + 2y) for ten, we obtain 2(4 + 2y) + 3y = 1, an equation with only one unknown.

The reason for this is that if 10 = 4 + 2y in one of the equations, then 10 must equal 4 + 2y in the other equation.

Step 3 Solve for the unknown.

Think, first remove parentheses.

Step 4 Substitute y = - 1 into either equation to detect the corresponding value for ten. Since nosotros have already solved the second equation for 10 in terms of y, we may utilize it.

We may substitute y = - 1 in either equation since y has the same value in both.

Thus, we have the solution (2,-1).

Recall, 10 is written first in the ordered pair.

Step v Cheque the solution in both equations. Call up that the solution for a system must be truthful for each equation in the system. Since

the solution (ii,-one) does bank check.

This checks: 2x + 3y = 1 and x - 2y = 4.

Check this ordered pair in both equations.
Neither of these equations had a variable with a coefficient of 1. In this case, solving by substitution is not the best method, just we will do information technology that way just to show information technology can be done. The next section will give united states of america an easier method.

SOLVING A Organisation OF LINEAR EQUATIONS By ADDITION

OBJECTIVES

Upon completing this section you should be able to solve a system of two linear equations by the addition method.

The addition method for solving a system of linear equations is based on two facts that we have used previously.

Kickoff we know that the solutions to an equation do not change if every term of that equation is multiplied by a nonzero number. 2d we know that if we add together the same or equal quantities to both sides of an equation, the results are still equal.

Instance one Solve past addition:

Note that nosotros could solve this organization by the substitution method, past solving the first equation for y. Solve this system past the exchange method and compare your solution with that obtained in this section.

Solution
Pace 1 Our purpose is to add the two equations and eliminate one of the unknowns so that we tin solve the resulting equation in one unknown. If nosotros add the equations as they are, nosotros will not eliminate an unknown. This means we must commencement multiply each side of one or both of the equations by a number or numbers that will atomic number 82 to the elimination of i of the unknowns when the equations are added.
Afterward carefully looking at the problem, we annotation that the easiest unknown to eliminate is y. This is done by starting time multiplying each side of the first equation by -2.

Note that each term must be multiplied past ( - 2).

Stride 2 Add the equations.

Step 3 Solve the resulting equation.

In this example we simply multiply each side past (-1).

Pace 4 Detect the value of the other unknown by substituting this value into 1 of the original equations. Using the offset equation,

Substitute ten = 4 in the second equation and run into if yous get the same value for y.

Step v If we check the ordered pair (four,-3) in both equations, we see that it is a solution of the system.

Example 2 Solve by addition:

Notation that in this organisation no variable has a coefficient of one. Therefore, the best method of solving information technology is the add-on method.

Solution
Step 1 Both equations will take to be inverse to eliminate one of the unknowns. Neither unknown volition be easier than the other, and then cull to eliminate either x or y.
To eliminate 10 multiply each side of the starting time equation by 3 and each side of the second equation past -2.

If you choose to eliminate y, multiply the beginning equation past - 2 and the second equation by 3. Exercise this and solve the system. Compare your solution with the one obtained in the case.

Step 2 Calculation the equations, nosotros obtain

Footstep three Solving for y yields

Step iv Using the get-go equation in the original system to find the value of the other unknown gives

Step five Bank check to meet that the ordered pair ( - 1,3) is a solution of the system.

The bank check is left upwardly to you.

STANDARD Grade

OBJECTIVES

Upon completing this section you should be able to:

Write a linear equation in standard form.
Solve a system of two linear equations if they are given in nonstandard form.

Equations in the preceding sections have all had no fractions, both unknowns on the left of the equation, and unknowns in the same guild.
Such equations are said to be in standard form. That is, they are in the form ax + past = c, where a, b and c are integers. Equations must be changed to the standard grade before solving past the addition method.

Example one Change 3x = five + 4y to standard form.

Solution 3x = 5 + 4y is not in standard form considering one unknown is on the correct. If nosotros add -4y to both sides, nosotros have 3x - 4y = 5, which is in standard form.

Be careful here. Many students forget to multiply the right side of the equation by 24.

Again, make sure each term is multiplied past 12.

Now add - 24x to both sides, giving - 24x + 9y = -10, which is in standard course. Usually, equations are written then the first term is positive. Thus we multiply each term of this equation by (- one).

Instead of saying "the first term is positive," we sometimes say "the leading coefficient is positive."

Discussion PROBLEMS WITH 2 UNKNOWNS

OBJECTIVES

Upon completing this section yous should be able to:

Decide when a word problem can be solved using 2 unknowns.
Make up one's mind the equations and solve the word problem.

Many word issues can be outlined and worked more than easily past using two unknowns.

Example 1 The sum of ii numbers is five. Three times the first number added to 5 times the second number is 9. Find the numbers.

Solution Let 10 = first number
y = 2nd number
The first statement gives usa the equation
ten + y = 5.
The second statement gives the states the equation
3x + 5 y = 9.
Nosotros now have the system

which we tin can solve by either method we have learned, to give
x = eight and y = - three.

Solve the organisation by commutation.

Example 2 Two workers receive a total of $136 for eight hours work. If one worker is paid $1.00 per hour more than the other, find the hourly charge per unit for each.

Solution Let x = hourly rate of one worker
y = hourly charge per unit of other worker.

Note that it is very of import to say what 10 and y represent.

The first statement gives us the equation
8x + 8y = 136.
The 2d statement gives the equation
ten = y + 1.
We now have the system (in standard form)

Solving gives ten = ix and y = 8. I worker's charge per unit is $9.00 per hour and the other's is $eight.00 per hr.

Solve this system by the add-on method.

SUMMARY

Key Words

The Cartesian coordinate organization is a method of naming points on a airplane.
Ordered pairs of numbers are used to designate points on a aeroplane.
A linear equation graphs a straight line.
The slope from one point on a line to another is the ratio .
The slope-intercept form of the equation of a line is y = mx + b.
A linear inequality graphs as a portion of the plane.
A system of ii linear equations consists of linear equations for which we wish to discover a simultaneous solution.
Contained equations accept unique solutions.
Inconsistent equations have no solution.
Dependent equations take infinitely many solutions.
A system of 2 linear inequalities consists of linear inequalities for which we wish to find a simultaneous solution.
The standard form of a linear equation is ax + by = c, where a, b, and c are existent numbers.

Procedures

To sketch the graph of a linear equation find ordered pairs of numbers that are solutions to the equation. Locate these points on the Cartesian coordinate organisation and connect them with a line.
To sketch the graph of a line using its slope:
Step i Write the equation of the line in the form y - mx + b.
Stride 2 Locate the j-intercept (0,b).
Step three Starting at (0,b), use the slope thou to locate a second point.
Footstep 4 Connect the two points with a direct line.
To graph a linear inequality:
Step one Replace the inequality symbol with an equal sign and graph the resulting line.
Footstep ii Check i point that is obviously in a particular half-aeroplane of that line to see if it is in the solution set of the inequality.
Step three If the point chosen is in the solution set up, then that entire one-half-plane is the solution set up. If the bespeak chosen is not in the solution set, then the other halfplane is the solution set.
To solve a system of 2 linear equations by graphing, graph the equations carefully on the aforementioned coordinate organization. Their point of intersection will be the solution of the system.
To solve a organisation of two linear inequalities by graphing, decide the region of the plane that satisfies both inequality statements.
To solve a system of two equations with two unknowns by substitution, solve for one unknown of one equation in terms of the other unknown and substitute this quantity into the other equation. So substitute the numerical value thus found into either equation to detect the value of the other unknown. Finally, check the solution in both equations.
To solve a system of ii equations with two unknowns by addition, multiply one or both equations by the necessary numbers such that when the equations are added together, one of the unknowns volition be eliminated. Solve for the remaining unknown and substitute this value into ane of the equations to find the other unknown. Cheque in both equations.
To solve a word problem with two unknowns find two equations that show a relationship betwixt the unknowns. Then solve the system. E'er check the solution in the stated problem.