The language of mathematics is particularly effective in representing relationshipsbetween two or more variables. As an example, let us consider the distance traveledin a certain length of time by a car moving at a constant speed of 40 miles per hour.We can represent this relationship by

- 1.A word sentence:

The distance traveled in miles is equal to forty times the number of hours traveled. - 2.An equation:

d = 40r. - 3.A tabulation of values.
- 4.A graph showing the relationship between time and distance.

We have already used word sentences and equations to describe such relationships;in this chapter, we will deal with tabular and graphical representations.

## 7.1SOLVING EQUATIONS IN TWO VARIABLES

ORDERED PAIRS

The equation d = 40f pairs a distance d for each time t. For example,

ift = 1,thend = 40

ift = 2,thend = 80

ift = 3,thend = 120

and so on.

The pair of numbers 1 and 40, considered together, is called a solution of theequation d = 40r because when we substitute 1 for t and 40 for d in the equation,we get a true statement. If we agree to refer to the paired numbers in a specifiedorder in which the first number refers to time and the second number refers todistance, we can abbreviate the above solutions as (1, 40), (2, 80), (3, 120), andso on. We call such pairs of numbers ordered pairs, and we refer to the first andsecond numbers in the pairs as components. With this agreement, solutions of theequation d - 40t are ordered pairs (t, d) whose components satisfy the equation.Some ordered pairs for t equal to 0, 1, 2, 3, 4, and 5 are

(0,0), (1,40), (2,80), (3,120), (4,160), and (5,200)

Such pairings are sometimes shown in one of the following tabular forms.

In any particular equation involving two variables, when we assign a value to oneof the variables, the value for the other variable is determined and thereforedependent on the first. It is convenient to speak of the variable associated with thefirst component of an ordered pair as the independent variable and the variableassociated with the second component of an ordered pair as the dependent variable. If the variables x and y are used in an equation, it is understood that replace-ments for x are first components and hence x is the independent variable andreplacements for y are second components and hence y is the dependent variable.For example, we can obtain pairings for equation

by substituting a particular value of one variable into Equation (1) and solving forthe other variable.

Example 1

Find the missing component so that the ordered pair is a solution to

2x + y = 4

a. (0,?)

b. (1,?)

c. (2,?)

Solution

ifx = 0,then2(0) + y = 4

y = 4

ifx = 1,then2(1) + y = 4

y = 2

ifx = 2,then2(2) + y = 4

y = 0

The three pairings can now be displayed as the three ordered pairs

(0,4),(1,2), and (2,0)

or in the tabular forms

EXPRESSING A VARIABLE EXPLICITLY

We can add -2x to both members of 2x + y = 4 to get

-2x + 2x + y = -2x + 4

y = -2x + 4

In Equation (2), where y is by itself, we say that y is expressed explicitly in termsof x. It is often easier to obtain solutions if equations are first expressed in such formbecause the dependent variable is expressed explicitly in terms of the independentvariable.

For example, in Equation (2) above,

ifx = 0,theny = -2(0)+ 4 = 4

ifx = 1,theny = -2(1)+ 4 = 2

ifx = 2 theny = -2(2)+ 4 = 0

We get the same pairings that we obtained using Equation (1)

(0,4),(1,2), and (2,0)

We obtained Equation (2) by adding the same quantity, -2x, to each memberof Equation (1), in that way getting y by itself. In general, we can write equivalentequations in two variables by using the properties we introduced in Chapter 3,where we solved first-degree equations in one variable.

Equations are equivalent if:

- The same quantity is added to or subtracted from equal quantities.
- Equal quantities are multiplied or divided by the same nonzero quantity.

Example 2

Solve 2y - 3x = 4 explicitly for y in terms of x and obtain solutions for x = 0,x = 1, and x = 2.

Solution

First, adding 3x to each member we get

2y - 3x + 3x = 4 + 3x

2y = 4 + 3x(continued)

Now, dividing each member by 2, we obtain

In this form, we obtain values of y for given values of x as follows:

In this case, three solutions are (0, 2), (1, 7/2), and (2, 5).

FUNCTION NOTATION

Sometimes, we use a special notation to name the second component of an orderedpair that is paired with a specified first component. The symbol f(x), which is oftenused to name an algebraic expression in the variable x, can also be used to denotethe value of the expression for specific values of x. For example, if

f(x) = -2x + 4

where f{x) is playing the same role as y in Equation (2) on page 285, then f(1)represents the value of the expression -2x + 4 when x is replaced by 1

f(l) = -2(1) + 4 = 2

Similarly,

f(0) = -2(0) + 4 = 4

and

f(2) = -2(2) + 4 = 0

The symbol f(x) is commonly referred to as function notation.

Example 3

If f(x) = -3x + 2, find f(-2) and f(2).

Solution

Replace x with -2 to obtain

f(-2) = -3(-2) + 2 = 8

Replace x with 2 to obtain

f(2) = -3(2) + 2 = -4

## 7.2GRAPHS OF ORDERED PAIRS

In Section 1.1, we saw that every number corresponds to a point in a line. Simi-larly, every ordered pair of numbers (x, y) corresponds to a point in a plane. Tograph an ordered pair of numbers, we begin by constructing a pair of perpendicularnumber lines, called axes. The horizontal axis is called the x-axis, the vertical axisis called the y-axis, and their point of intersection is called the origin. These axesdivide the plane into four quadrants, as shown in Figure 7.1.

Now we can assign an ordered pair of numbers to a point in the plane by referringto the perpendicular distance of the point from each of the axes. If the firstcomponent is positive, the point lies to the right of the vertical axis; if negative, itlies to the left. If the second component is positive, the point lies above thehorizontal axis; if negative, it lies below.

Example 1

Graph (3, 2), (-3, 2), (-3, -2), and (3, -2) on a rectangular coordinate system.

Solution

The graph of (3, 2) lies 3 units to the right ofthe y-axis and 2 units above the x-axis;the graph of (-3,2) lies 3 units to the left of they-axis and 2 units above the x-axis;the graph of (-3, -2) lies 3 units to the left ofthe y-axis and 2 units below the x-axis;the graph of (3, -2) lies 3 units to the right ofthe y-axis and 2 units below the x-axis.

The distance y that the point is located from the x-axis is called the ordinateof the point, and the distance x that the point is located from the y-axis is calledthe abscissa of the point. The abscissa and ordinate together are called the rectan-gular or Cartesian coordinates of the point (see Figure 7.2).

## 7.3GRAPHING FIRST-DEGREE EQUATIONS

In Section 7.1, we saw that a solution of an equation in two variables is an orderedpair. In Section 7.2, we saw that the components of an ordered pair are thecoordinates of a point in a plane. Thus, to graph an equation in two variables, wegraph the set of ordered pairs that are solutions to the equation. For example, wecan find some solutions to the first-degree equation

y = x + 2

by letting x equal 0, -3, -2, and 3. Then,

forx = 0,y=0+2=2

forx = 0,y = -3 + 2 = -1

forx = -2,y = -2 + 2 - 0

forx = 3,y = 3 + 2 = 5

and we obtain the solutions

(0,2), (-3,-1), (-2,0), and (3,5)

which can be displayed in a tabular form as shown below.

If we graph the points determined by theseordered pairs and pass a straight line throughthem, we obtain the graph of all solutions ofy = x + 2, as shown in Figure 7.3. That is,every solution of y = x + 2 lies on the line,and every point on the line is a solution ofy = x + 2.

The graphs of first-degree equations in twovariables are always straight lines; therefore,such equations are also referred to as linearequations.

In the above example, the values we used forx were chosen at random; we could have usedany values of x to find solutions to the equation.The graphs of any other ordered pairs that are solutions of the equation would alsobe on the line shown in Figure 7.3. In fact, each linear equation in two variableshas an infinite number of solutions whose graph lies on a line. However, we onlyneed to find two solutions because only two points are necessary to determine astraight line. A third point can be obtained as a check.

To graph a first-degree equation:

- Construct a set of rectangular axes showing the scale and the variable repre-sented by each axis.
- Find two ordered pairs that are solutions of the equation to be graphed byassigning any convenient value to one variable and determining the corre-sponding value of the other variable.
- Graph these ordered pairs.
- Draw a straight line through the points.
- Check by graphing a third ordered pair that is a solution of the equation andverify that it lies on the line.

Example 1

Graph the equation y = 2x - 6.

Solution

We first select any two values of x to find the associated values of y.

We will use 1 and 4 for x.

If x = 1, y = 2(1) - 6 = -4

if x = 4, y = 2(4) - 6 = 2

Thus, two solutions of the equation are

(1, -4) and (4, 2).

Next, we graph these ordered pairs and draw a straight line through the points as shownin the figure. We use arrowheads to show thatthe line extends infinitely far in both directions.Any third ordered pair that satisfies theequation can be used as a check:

if x = 5, y = 2(5) -6 = 4

We then note that the graph of (5, 4) also lies on the line

To find solutions to an equation, as we have noted it is often easiest to first solveexplicitly for y in terms of x.

Example 2

Graph x + 2y = 4.

Solution

We first solve for y in terms of x to get

We now select any two values of x to find the associated values of y. We will use2 and 0 for x.

Thus, two solutions of the equation are (2, 1) and (0, 2).

Next, we graph these ordered pairs andpass a straight line through the points, asshown in the figure.

Any third ordered pair that satisfies theequation can be used as a check:

We then note that the graph of (-2, 3) alsolies on the line.

SPECIAL CASES OF LINEAR EQUATIONS

The equation y = 2 can be written as

0x + y = 2

and can be considered a linear equation in twovariables where the coefficient of x is 0. Somesolutions of 0x + y = 2 are

(1,2), (-1,2), and (4,2)

In fact, any ordered pair of the form (x, 2) isa solution of (1). Graphing the solutionsyields a horizontal line as shown in Figure7.4.

Similarly, an equation such as x = -3 canbe written as

x + 0y = -3

and can be considered a linear equation in twovariables where the coefficient of y is 0.

Some solutions of x + 0y = -3 are(-3, 5), (-3, 1), and (-3, -2). In fact, anyordered pair of the form (-3, y) is a solutionof (2). Graphing the solutions yields a verticalline as shown in Figure 7.5.

Example 3

Graph

a. y = 3

b. x=2

Solution

a. We may write y = 3 as Ox + y =3.

Some solutions are (1, 3), (2,3), and (5, 3).

b. We may write x = 2 as x + Oy = 2.

Some solutions are (2, 4), (2, 1), and (2, -2).

## 7.4 INTERCEPT METHOD OF GRAPHING

In Section 7.3, we assigned values to x in equations in two variables to find thecorresponding values of y. The solutions of an equation in two variables that aregenerally easiest to find are those in which either the first or second component is0. For example, if we substitute 0 for x in the equation

3x + 4y = 12

we have

3(0) + 4y = 12

y = 3

Thus, a solution of Equation (1) is (0, 3). We can also find ordered pairs that aresolutions of equations in two variables by assigning values to y and determining thecorresponding values of x. In particular, if we substitute 0 for y in Equation (1), weget

3x + 4(0) = 12

x = 4

and a second solution of the equation is (4, 0). We can now use the ordered pairs(0, 3) and (4, 0) to graph Equation (1). The graph is shown in Figure 7.6. Noticethat the line crosses the x-axis at 4 and the y-axis at 3. For this reason, the number4 is called the x-intercept of the graph, and the number 3 is called the y-intercept.

This method of drawing the graph of a linear equation is called the interceptmethod of graphing. Note that when we use this method of graphing a linearequation, there is no advantage in first expressing y explicitly in terms of x.

Example 1

Graph 2x - y = 6 by the intercept method.

Solution

We find the x-intercept by substituting 0 for y in the equation to obtain

2x - (0) = 6

2x = 6

x = 3

Now, we find the y-intercept by substitutingfor x in the equation to get

2(0) - y = 6

-y = 6

y = -6

The ordered pairs (3, 0) and (0, -6) are solutions of 2x - y = 6. Graphing thesepoints and connecting them with a straight line give us the graph of 2x - y = 6.If the graph intersects the axes at or near the origin, the intercept method is notsatisfactory. We must then graph an ordered pair that is a solution of the equationand whose graph is not the origin or is not too close to the origin.

Example 2

Graph y = 3x.

Solution

We can substitute 0 for x and find

y = 3(0) = 0

Similarly, substituting 0 for y, we get

0 = 3.x, x = 0

Thus, 0 is both the x-intercept and the y-intercept.

Since one point is not sufficient to graphy = 3x, we resort to the methods outlined inSection 7.3. Choosing any other value for x,say 2, we get

y = 3(2) = 6

Thus, (0, 0) and (2, 6) are solutions to theequation. The graph of y = 3x is shown at theright.

## 7.5SLOPE OF A LINE

SLOPE FORMULA

In this section, we will study an important property of a line. We will assign anumber to a line, which we call slope, that will give us a measure of the "steepness"or "direction" of the line.

It is often convenient to use a special notation to distinguish between the rectan-gular coordinates of two different points. We can designate one pair of coordinatesby (x_{1}, y_{1} (read "x sub one, y sub one"), associated with a point P_{1}, and a secondpair of coordinates by (x_{2}, y_{2}), associated with a second point P_{2}, as shown in Figure7.7. Note in Figure 7.7 that when going from P_{1} to P_{2}, the vertical change (orvertical distance) between the two points is y_{2} - y_{1} and the horizontal change (orhorizontal distance) is x_{2} - x_{1}.

The ratio of the vertical change to the horizontal change is called the slope of theline containing the points P_{1} and P_{2}. This ratio is usually designated by m. Thus,

Example 1

Find the slope of the line containing the twopoints with coordinates (-4, 2) and (3, 5) asshown in the figure at the right.

Solution

We designate (3, 5) as (x_{2}, y_{2}) and (-4, 2)as (x_{1}, y_{1}). Substituting into Equation (1)yields

Note that we get the same result if we subsitute -4 and 2 for x_{2} and y_{2} and 3 and5 for x_{1} and y_{1}

Lines with various slopes are shown in Figure 7.8 below. Slopes of the lines thatgo up to the right are positive (Figure 7.8a) and the slopes of lines that go downto the right are negative (Figure 7.8b). And note (Figure 7.8c) that because allpoints on a horizontal line have the same y value, y_{2} - y_{1} equals zero for any twopoints and the slope of the line is simply

Also note (Figure 7.8c) that since all points on a vertical have the same x value,x_{2} - x_{1} equals zero for any two points. However,

is undefined, so that a vertical line does not have a slope.

PARALLEL AND PERPENDICULAR LINES

Consider the lines shown in Figure 7.9. Line l_{1} has slope m_{1} = 3, and line l_{2} hasslope m_{2} = 3. In this case,

These lines will never intersect and are called parallel lines. Now consider the linesshown in Figure 7.10. Line l_{1}, has slope m_{1} = 1/2 and line l_{2} has slope m_{2} = -2.In this case,

These lines meet to form a right angle and are called perpendicular lines.

In general, if two lines have slopes and m2:

- a.The lines are parallel if they have the same slope, that is,if m

_{1}= m

_{2}.

b.The lines are perpendicular If the product of their slopesis -1, that is, if m

_{1}* m

_{2}= -1.

## 7.6EQUATIONS OF STRAIGHT LINES

POINT-SLOPE FORM

In Section 7.5, we found the slope of a straight line by using the formula

Let us say we know that a line goes through the point (2, 3) and has a slope of 2.If we denote any other point on the line as P(x, y) (See Figure 7.1 la), by the slopeformula

Thus, Equation (1) is the equation of the line that goes through the point (2, 3) andhas a slope of 2.

In general let us say we know a line passes through a point P_{1}(x_{1}, y_{1} and hasslope m. If we denote any other point on the line as P(x, y) (see Figure 7.11 b), bythe slope formula

Equation (2) is called the point-slope form for a linear equation. In Equation (2),m, x_{1} and y_{1} are known and x and y are variables that represent the coordinates ofany point on the line. Thus, whenever we know the slope of a line and a point onthe line, we can find the equation of the line by using Equation (2).

Example 1

A line has slope -2 and passes through point (2, 4). Find the equation of the line.

Solution

Substitute -2 for m and (2, 4) for (x_{1}, y_{1}) in Equation (2)

Thus, a line with slope -2 that passes through the point (2, 4) has the equationy = -2x + 8. We could also write the equation in equivalent forms y + 2x = 8,2x + y = 8, or 2x + y - 8 = 0.

SLOPE-INTERCEPT FORM

Now consider the equation of a line with slope m and y-intercept b as shown inFigure 7.12. Substituting 0 for x_{1} and b for y_{1} in the point-slope form of a linearequation, we have

y - b = m(x - 0)

y - b = mx

or

y = mx + b

Equation (3) is called the slope-intercept formfor a linear equation. The slope and y-interceptcan be obtained directly from an equation inthis form.

Example 2 If a line has the equation

then the slope of the line must be -2 and the y-intercept must be 8. Similarly, thegraph of

y = -3x + 4

has a slope -3 and a y-intercept 4; and the graph of

has a slope 1/4 and a y-intercept -2.

If an equation is not written in x = mx + b form and we want to know the slopeand/or the y-intercept, we rewrite the equation by solving for y in terms of x.

Example 3

Find the slope and y-intercept of 2x - 3y = 6.

Solution

We first solve for y in terms of x by adding -2x to each member.

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

- 3y = 6 - 2x

Now dividing each member by -3, we have

Comparing this equation with the form y = mx + b, we note that the slope m (thecoefficient of x) equals 2/3, and the y-intercept equals -2.

## 7.7DIRECT VARIATION

A special case of a first-degree equation in two variables is given by

y = kx (k is a constant)

Such a relationship is called a direct variation. We say that the variable y variesdirectly as x.

Example 1

We know that the pressure P in a liquid varies directly as the depth d below thesurface of the liquid. We can state this relationship in symbols as

P = kd

In a direct variation, if we know a set of conditions on the two variables, and ifwe further know another value for one of the variables, we can find the value ofthe second variable for this new set of conditions.

In the above example, we can solve for the constant k to obtain

Since the ratio P/d is constant for each set of conditions, we can use a proportionto solve problems involving direct variation.

Example 2

If pressure P varies directly as depth d, and P = 40 when d = 10, find P whend = 15.

Solution

Since the ratio P/d is constant, we can substitute values for P and d and obtain theproportion

Thus, P = 60 when d = 15.

## 7.8INEQUALITIES IN TWO VARIABLES

In Sections 7.3 and 7.4, we graphed equations in two variables. In this section wewill graph inequalities in two variables. For example, consider the inequality

y ≤ -x + 6

The solutions are ordered pairs of numbers that "satisfy" the inequality. That is,(a, b) is a solution of the inequality if the inequality is a true statement after wesubstitute a for x and b for y.

Example 1

Determine if the given ordered pair is a solution of y = -x + 6.

a. (1, 1)

b. (2, 5)

Solution

The ordered pair (1, 1) is a solution because, when 1 is substituted for x and 1 issubstituted for y, we get

(1) = -(1) + 6, or 1 = 5

which is a true statement. On the other hand, (2, 5) is not a solution because when2 is substituted for x and 5 is substituted for y, we obtain

(5)= -(2) + 6, or 5 = 4

which is a false statement.

To graph the inequality y = -x + 6, we first graph the equation y = -x + 6shown in Figure 7.13. Notice that (3, 3), (3, 2), (3, 1), (3, 0), and so on, associatedwith the points that are on or below the line, are all solutions of the inequalityy = -x + 6, whereas (3,4), (3, 5), and (3,6), associated with points above theline are not solutions of the inequality. In fact, all ordered pairs associated withpoints on or below the line are solutions of y = - x + 6. Thus, every point on orbelow the line is in the graph. We represent this by shading the region below theline (see Figure 7.14).

In general, to graph a first-degree inequality in two variables of the formAx + By = C or Ax + By = C, we first graph the equation Ax + By = C andthen determine which half-plane (a region above or below the line) contains thesolutions. We then shade this half-plane. We can always determine which half-plane to shade by selecting a point (not on the line of the equation Ax + By = C)and testing to see if the ordered pair associated with the point is a solution of thegiven inequality. If so, we shade the half-plane containing the test point; otherwise,we shade the other half-plane. Often, (0, 0) is a convenient test point.

Example 2

Graph 2x+3y = 6

Solution

We first graph the line 2x + 3y = 6 (see graph a). Using the origin as a test point,we determine whether (0, 0) is a solution of 2x + 3y ≥ 6. Since the statement

2(0) + 3(0) = 6

is false, (0, 0) is not a solution and we shade the half-plane that does not containthe origin (see graph b).

When the line Ax + By = C passes through the origin, (0, 0) is not a valid testpoint since it is on the line.

Example 3

Graph y = 2x.

Solution

We begin by graphing the line y = 2x (see graph a). Since the line passes throughthe origin, we must choose another point not on the line as our test point. We willuse (0, 1). Since the statement

(1) = 2(0)

is true, (0, 1) is a solution and we shade the half-plane that contains (0, 1) (seegraph b).

If the inequality symbol is '< or > , the points on the graph of Ax + By = Care not solutions of the inequality. We then use a dashed line for the graph ofAx + By = C.

## CHAPTER SUMMARY

A solution of an equation in two variables is an ordered pair of numbers. In theordered pair (x, y), x is called the first component and y is called the secondcomponent. For an equation in two variables, the variable associated with the firstcomponent of a solution is called the independent variable and the variableassociated with the second component is called the dependent variable.Function notation f(x) is used to name an algebraic expression in x. When x inthe symbol f(x) is replaced by a particular value, the symbol represents the valueof the expression for that value of x.

The intersection of the two perpendicular axes in a coordinate systemis called theorigin of the system, and each of the four regions into which the plane is dividedis called a quadrant. The components of an ordered pair (x, y) associated with apoint in the plane are called the coordinates of the point; x is called the abscissaof the point and y is called the ordinate of the point.

The graph of a first-degree equation in two variables is a straight line. That is, everyordered pair that is a solution of the equation has a graph that lies in a line, andevery point in the line is associated with an ordered pair that is a solution of theequation.

The graphs of any two solutions of an equation in two variables can be used toobtain the graph of the equation. However, the two solutions of an equation in twovariables that are generally easiest to find are those in which either the first orsecond component is 0. The x-coordinate of the point where a line crosses the x-axisis called the x-intercept of the line, and the y-coordinate of the point where a linecrosses the y-axis is called they-intercept of the line. Using the intercepts to graphan equation is called the intercept method of graphing.

The slope of a line containing the points P

_{1}(x_{1}, y_{1}) and P_{2}(x_{2}, y_{2}) is given byTwo lines are parallel if they have the same slope (m

_{1}= m_{2}).Two lines are perpendicular if the product of their slopes is - l(m

_{1}* m_{2}= -1).The point-slope form of a line with slope m and passing through the point (x

_{1}, y_{1})isy - y

_{1}- m(x - x_{1})The slope-intercept form of a line with slope m and y-intercept b is

y = mx + b

A relationship determined by an equation of the form

y = kx (k a constant)

is called a direct variation.

A solution of an inequality in two variables is an ordered pair of numbers that,when substituted into the inequality, makes the inequality a true statement. Thegraph of a linear inequality in two variables is a half-plane.The symbols introduced in this chapter appear on the inside front covers.