# Graphing equations with step by step math problem solver (2023)

The language of mathematics is particularly effective at showing relationships between two or more variables. As an example, consider the distance traveled by a car traveling at a constant speed of 40 miles per hour over a period of time. We can represent this relationship by

1. 1. One word sentence:
The distance traveled in miles is equal to forty times the hours traveled.
2. 2. An equation:
d = 40r.
3. 3. A tabular list of values.
4. 4. A graph showing the relationship between time and distance.

We have already used phrases and equations to describe such relationships; In this chapter we will deal with tabular and graphical representations.

## 7.1 SOLVE EQUATIONS IN TWO VARIABLES

ORDERED PAIRS

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

ift = 1, then end = 40
ift = 2, then end = 80
if = 3, then = 120

etc.

The pair of numbers 1 and 40 together is called the solution of the equation d = 40r, because if we replace t with 1 and d with 40 in the equation, we get a true statement. If we agree to refer to the paired numbers in a specific order, where the first number refers to time and the second number refers to distance, we can use the above solutions like (1, 40), (2, 80), (3 , 120). ), etc. We call these pairs numbers of ordered pairs and refer to the first and second numbers of the pairs as components. With this convention, the solutions of the equation d - 40t are ordered pairs (t,d) whose components satisfy the equation. Some ordered pairs for t are 0, 1, 2, 3, 4, and 5.

(0.0), (1.40), (2.80), (3.120), (4.160) y (5.200)

These pairings are sometimes presented in one of the following tabular forms.

In a given equation of two variables, when we assign a value to one of the variables, the value of the other variable is determined and therefore depends on the first. It is convenient to call the variable associated with the first component of an ordered pair the independent variable and the variable associated with the second component of an ordered pair the dependent variable. When the variables x and y are used in an equation, it is understood that substitutions for x are the first components, and therefore x is the independent variable, and substitutions for y are the second components, and therefore y is the independent variable. dependent variable. For example, we can get pairs for the equation

Substituting a specific value of one variable into equation (1) and solving for the other variable.

Example 1

Find the missing component for the ordered pair for which you want to find a solution

2x + y = 4

R. (0,?)

B. (1,?)

C. (2,?)

Solution

se x = 0,então2(0) + y = 4
y = 4

se x = 1,então2(1) + y = 4
y = 2

se x = 2,então2(2) + y = 4
y = 0

The three pairs can now be viewed as the three ordered pairs

(0.4), (1.2) y (2.0)

or in tabular form

TO EXPRESS A VARIABLE EXPRESSLY

We can add -2x to either side of 2x + y = 4 to get

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

In equation (2), where y stands alone, we say that y is expressed explicitly by x. Usually, it is easier to obtain solutions if the equations are first expressed in this way, since the dependent variable is expressed explicitly by the independent variable.

For example, in equation (2) above,

six = 0,so = -2(0)+ 4 = 4
six = 1, so = -2(1)+ 4 = 2
six = 2 then = -2(2)+ 4 = 0

We obtain the same pairs as with equation (1)

(0.4), (1.2) y (2.0)

We obtain Equation (2) by adding the same quantity, -2x, to each member of Equation (1), giving us the same y. In general, we can write equivalent equations in two variables using the properties we introduced in Chapter 3, where we solved first degree equations in one variable.

The equations are equivalent if:

1. The same amount is added to or subtracted from equal amounts.
2. Equal amounts are multiplied or divided by the same non-zero amount.

example 2

Explicitly solve 2y - 3x = 4 for y for x and get solutions for x = 0, x = 1, and x = 2.

Solution
First, if we add 3x to each member, we get

2y - 3x + 3x = 4 + 3x
2y = 4 + 3x (continued)

Now, if we divide each member by 2, we get

In this way, we obtain y-values ​​for given x-values ​​as follows:

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

FUNCTIONAL DESIGNATION

We sometimes use a special notation to name the second component of an ordered pair paired with a particular first component. The symbol f(x), which is often used to denote an algebraic expression about the variable x, can also be used to denote the value of the expression for specific values ​​of x. for example when

f(x) = -2x + 4

where f{x) plays the same role as y in Equation (2) on page 275, so f(1) represents the value of the expression –2x + 4 when x is replaced by 1

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

Similar,

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

(Video) Solving Systems of Equations By Graphing

mi

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

The symbol f(x) is commonly known as a function notation.

Example 3

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

Solution

Replace x with -2 to get
f(-2) = -3(-2) + 2 = 8

Replace x with 2 to get
f(2) = -3(2) + 2 = -4

## 7.2 GRAPHICS OF ORDERED PAIRS

In Section 1.1 we saw that each number corresponds to a point on a line. Likewise, each ordered pair of numbers (x,y) corresponds to a point in a plane. To represent an ordered pair of numbers, we first construct two number lines at right angles, called axes. The horizontal axis is called the x axis, the vertical axis is called the y axis, and their point of intersection is called the origin. These axes divide the plane into four quadrants as shown in Figure 7.1.

We can now assign an ordered pair of numbers to a point in the plane by referring to the point's perpendicular distance from each of the axes. If the first component is positive, the point is to the right of the vertical axis; if it is negative, it is to the left. If the second component is positive, the point is above the horizontal axis; if it's negative, stay down.

Example 1

Draw (3, 2), (-3, 2), (-3, -2), and (3, -2) in a rectangular coordinate system.

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

The distance y that the point is from the x axis is called the ordinate of the point, and the distance x that the point is from the y axis is called the abscissa of the point. The abscissa and the ordinate together are called the Cartesian or rectangular coordinates of the point (see Figure 7.2).

## 7.3 DIAGRAM OF FIRST DEGREE EQUATIONS

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

y = x + 2

where x remains equal to 0, -3, -2, and 3. So,

for x = 0,y=0+2=2
for x = 0, y = -3 + 2 = -1
for x = -2, y = -2 + 2 -
for x = 3, y = 3 + 2 =

and we get the solutions

(0.2), (-3,-1), (-2.0) y (3.5)

which can be seen in tabular form as shown below.

If we plot the points determined by these ordered pairs and draw a straight line through them, we obtain the graph of all solutions of y = 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 of y = x + 2.

The graphs of first degree equations in two variables are always straight lines; therefore, such equations are also called linear equations.

In the example above, the values ​​we used for x were chosen at random; We could have used any value of x to find solutions to the equation. The graphs of all other ordered pairs that are solutions of the equation would also lie on the line shown in Figure 7.3. In fact, every linear equation in two variables has infinitely many solutions whose graph lies on a line. However, we only need to find two solutions because it only takes two points to find a line. A third point can be reached as verification.

To graph a first degree equation:

1. Construct a set of rectangular axes showing the scale and the variable represented by each axis.
2. Find two ordered pairs that are solutions to the equation to be represented by assigning an appropriate value to one variable and finding the corresponding value of the other variable.
3. Draw these ordered pairs.
4. Draw a straight line through the points.
5. Check this by graphing a third-order pair that is a solution to the equation, and verify that it lies on the line.

Example 1

Draw the equation y = 2x - 6.

Solution
We first choose any two x-values ​​to find the associated y-values.
We use 1 and 4 for x.
When x = 1, y = 2(1) - 6 = -4
when x = 4, y = 2(4) - 6 = 2
So there are two solutions to the equation.
(1, -4) y (4, 2).
We then plot these ordered pairs and draw a straight line through the points as shown in the figure. We use arrowheads to show that the line extends infinitely in any direction. Any third-order pair that satisfies the equation can be used as a check:
when x = 5, y = 2(5) -6 = 4
Then we notice that the graph of (5, 4) also lies on the line
As we've noted, to find solutions to an equation, it's usually easiest to first solve explicitly for y with respect to x.

example 2

Graph x + 2y = 4.

Solution
First we solve for y with respect to x to get

Now we select any two x-values ​​to find the associated y-values. We use 2 and 0 for x.

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

We then plot these ordered pairs and draw a straight line through the points as shown in the figure.

Any third-order pair that satisfies the equation can be used as a check:

Then we notice that the graph of (-2, 3) also lies on the line.

SPECIAL CASES OF LINEAR EQUATIONS

The equation y = 2 can be written as

0x + y = 2

y can be viewed as a linear equation in two variables, where the coefficient of x is 0. Some solutions of 0x + y = 2 are

(1.2), (-1.2) y (4.2)

In fact, every ordered pair of the form (x, 2) is a solution of (1). Graphing the solutions creates a horizontal line as shown in Figure 7.4.

(Video) MalMath: Step by step math problem solver OVERVIEW

Similarly, an equation such as x = -3 can be written as

x + 0y = -3

y can be viewed as a linear equation in two variables, where the coefficient of y is 0.

Some solutions of x + 0y = -3 are (-3, 5), (-3, 1), and (-3, -2). In fact, every ordered pair of the form (-3, y) is a solution of (2). Graphing the solutions creates a vertical line as shown in Figure 7.5.

Example 3

Chart

for. y = 3
B x=2

Solution
A. We can write y = 3 as Ox + y =3.
Some solutions are (1, 3), (2,3), and (5, 3).

B. We can write x = 2 as x + Oy = 2.
Some solutions are (2, 4), (2, 1), and (2, -2).

## 7.4 INTERZEPTION CHART METHOD

In Section 7.3, we assigned values ​​to x in equations in two variables to find the corresponding values ​​of y. The two-variable solutions to an equation that are usually easiest to find are those in which the first or second component is 0. For example, if we replace x with 0 in the equation

3x + 4y = 12

Have

3(0) + 4a = 12
y = 3

Therefore, a solution of equation (1) is (0, 3). We can also find ordered pairs that solve equations in two variables by assigning values ​​to y and finding the corresponding values ​​of x. In particular, if we replace y in equation (1) by 0, we get

3x + 4(0) = 12
x = 4

and a second solution of the equation is (4, 0). Now we can use the ordered pairs (0, 3) and (4, 0) to graph equation (1). The diagram is shown in Figure 7.6. Notice that the line intersects the x-axis at 4 and the y-axis at 3. For this reason, the number 4 is called the x-intercept of the graph and the number 3 is called the y-intercept.

This method of graphing a linear equation is called the graph-intercept method. Note that when using this method to graph a linear equation, there is no benefit to first explicitly expressing y in terms of x.

Example 1

Draw 2x - y = 6 by the intersection method.

Solution
We find the x-intercept by substituting 0 for y in the resulting equation

2x - (0) = 6
2x = 6
x = 3

Now we find the y-intercept by substituting x into the equation that gets

2(0) - y = 6
-y = 6
y = -6

The ordered pairs (3, 0) and (0, -6) are solutions of 2x - y = 6. Graphing these points and connecting them with a straight line gives the graph of 2x - y = 6. If the graph intersects the axes at or near the origin, the interception method is not satisfactory. So we need to draw an ordered pair that is a solution to the equation and whose graph is not the origin or not very close to the origin.

example 2

Graph y = 3x.

Solution
We can replace x with 0 and find
y = 3(0) = 0
Similarly, if we replace y with 0, we get
0 = 3.x, x = 0
So 0 is the x-intercept and y-intercept.

Since one point is not enough to write = 3x, we use the methods described in Section 7.3. If we choose a different value for x, say 2, we get

y = 3(2) = 6

So (0, 0) and (2, 6) are solutions of the equation. The graph of y = 3x is shown on the right.

## 7.5 SLOPE OF A LINE

SLOPE FORMULA

In this section we will examine an important property of a line. We assign a number to a line, which we call slope, which gives us a measure of the "slope" or "direction" of the line.

It is often convenient to use a special notation to distinguish between the orthogonal coordinates of two different points. We can use a pair of coordinates with (x1, Of1(read "x sub one, y sub one"), connected with a point P1, and a second pair of coordinates by (x2, Of2), connected to a second point P2, as shown in Figure 7.7. Notice in Figure 7.7 that by changing from P1Principal2, is the vertical change (ortho distance) between the two points and2-y1and the horizontal offset (or horizontal distance) is x2- X1.

The ratio of the vertical change to the horizontal change is called the slope of the line containing the points P1mi pag2. This relationship is commonly denoted by m. For this reason,

Example 1

Find the slope of the line that contains the two points with coordinates (-4, 2) and (3, 5) as shown in the figure to the right.

Solution
We denote (3, 5) as (x2, Of2) and (-4, 2) as (x1, Of1). Substituting into equation (1) gives

Note that if we replace -4 and 2 with x, we get the same result2And you2y 3 y 5 for x1And you1

Lines with different slopes are shown in Figure 7.8 below. The slopes of the lines that rise to the right are positive (Figure 7.8a) and the slopes of the lines that fall to the right are negative (Figure 7.8b). And notice (Figure 7.8c) that since all points on a horizontal line have the same y value, y2-y1is zero for any two points and the slope of the line is simple

Also note (Figure 7.8c) that since all points on a vertical have the same value of x, x2- X1is zero for any two points. However,

(Video) Solving Quadratic Equations Graphically - Corbettmaths

is undefined, so a vertical line has no slope.

PARALLEL AND RECTANGULAR LINES

Look at the lines in Figure 7.9. line l1has pending m1= 3 and line l2Hasshang m2= 3. In this case,

These lines will never cross and are called parallel lines. Now consider the lines shown in Figure 7.10. line l1, has slope m1= 1/2 and line l2has pending m2= -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:

El. Lines are parallel if they have the same slope, that is, if m1= metro2.
B. Lines are perpendicular if the product of their slopes is -1, that is, if m1* M2= -1.

## 7.6 EQUATIONS OF STRAIGHT LINES

POINT EARRING SHAPE

In Section 7.5 we determined the slope of a straight line using the formula

Suppose we know that a line passes 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 7.11a), we use the formula for the earring

Then Equation (1) is the equation of the line that passes through the point (2,3) and has slope 2.

In general, suppose we know that a line passes through a point P1(X1, Of1y has slope m If we denote any other point on the line as P(x, y) (see Figure 7.11b), by the slope formula

Equation (2) is called the point-slope form of a linear equation. In equation (2) m, x1And you1are known and x and y are variables that represent the coordinates of any point on the line. So, as long as we know the slope of a line and a point on the line, we can find the equation of the line using Equation (2).

Example 1

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

Solution
Replace -2 with m and (2, 4) with (x1, Of1) in equation (2)

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

TILT-INTERCEPTION-FORMULAR

Now consider the equation of a straight line with slope m and axis-intercept b, as shown in Figure 7.12. Replace 0 with x1e b for y1In point-slope form of a linear equation we have

y - b = m(x - 0)
y - b = mx

o

y = meter x + second

Equation (3) is known as the slope-intercept form for a linear equation. The slope and y-intercept can be obtained directly from an equation in this form.

Example 2 If a line has the equation

so the slope of the line must be -2 and the y-intercept must be 8. Likewise, the graph of

y = -3x + 4

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

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

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

Example 3

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

Solution
First we solve for y with respect to x by adding -2x to each side.

2x - 3a - 2x = 6 - 2x
- 3 years = 6 - 2x

Now, if we divide each member by -3, we have

Comparing this equation to the form y = mx + b, we find that the slope m (the coefficient of x) is 2/3 and the y-intercept is -2.

## 7.7 DIRECT 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 direct variation. We say that the variable y varies directly with x.

Example 1

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

P = kd

(Video) Graphing a linear equation by rewriting from standard form to slope intercept form

In a direct variation, if we know one set of conditions for the two variables, and if we know a different value for one of the variables, we can find the value of the second variable for that new set of conditions.

In the above example, we can solve to get the constant k

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

example 2

If the pressure P varies directly with depth d and P = 40 at d = 10, you will find P at d = 15.

Solution
Since the ratio P/d is constant, we can plug in the values ​​of P and d and obtain the ratio

So P = 60 when d = 15.

## 7.8 INEQUALITIES IN TWO VARIABLES

In Sections 7.3 and 7.4 we drew equations in two variables. In this section we will 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 replacing a by x and b by y.

Example 1

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

R. (1, 1)
B. (2, 5)

Solution
The ordered pair (1, 1) is a solution because replacing x with 1 and y with 1 we get

(1) = -(1) + 6 o 1 = 5

which is a true statement. On the other hand, (2, 5) is not a solution, because replacing x by 2 and y by 5 we get

(5)= -(2) + 6 o 5 = 4

which is a false statement.

To graph the inequality y = -x + 6, we first graph the equation y = -x + 6 shown in Figure 7.13. Notice that (3, 3), (3, 2), (3, 1), (3 , 0), etc associated with points that lie on or below the line, all solutions of the inequality = -x + are 6, while (3,4), (3,5) and (3,6), connected to points by above the line, are not solutions of the inequality. In fact, all ordered pairs associated with points on or below the line are solutions of y = - x + 6. Therefore, every point is on or below the line on the graph. We represent this by shading the area below the line (see Figure 7.14).

In general, to graph a first degree inequality in two variables of the form Ax + By = C or Ax + By = C, we first sketch the equation Ax + By = C and then determine which half-plane (an area above or below) below of the line ) contains the solutions. Then we shade this middle plane. We can always determine which half-plane to shade by choosing a point (not on the line of the equation Ax + By = C) and testing whether the ordered pair associated with the point is a solution of the given inequality. In this case we shade the half plane that contains the test point; Otherwise we shade the other half plane. Often (0, 0) is a convenient test point.

example 2

Gráfico 2x + 3y = 6

Solution
First we draw 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. as a statement

2(0) + 3(0) = 6

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

If the line Ax + By = C passes through the origin, then (0, 0) is not a valid test point because it lies on the line.

Example 3

Graph y = 2x.

Solution
We start by drawing the line y = 2x (see graph a). Since the line passes through the origin, we must choose a different point that does not lie on the line as our test point. We will use (0, 1). from the statement

(1) = 2(0)

is true, (0, 1) is a solution, and we shade the half-plane containing (0, 1) (see graph b).

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

## CHAPTER SUMMARY

1. A solution to an equation in two variables is an ordered pair of numbers. In the ordered pair (x,y), x is called the first component and y is called the second component. In a two-variable equation, the variable associated with the first component of a solution is called the independent variable, and the variable associated with the second component is called the dependent variable. The function notation f(x) is used to denote an algebraic expression in x. If x in the symbol f(x) is replaced by a specific value, the symbol represents the value of the expression for that value of x.

2. The intersection of two orthogonal axes in a coordinate system is called the origin of the system, and each of the four areas into which the plane is divided is called a quadrant. The components of an ordered pair (x, y) associated with a point in the plane are called the coordinates of the point; x is called the abscissa of the point and y is called the ordinate of the point.

3. The graph of a first degree equation in two variables is a straight line. That is, each ordered pair that is a solution of the equation has a graph that lies on a line, and each point on the line is associated with an ordered pair that is a solution of the equation.

The graphs of any two solutions to an equation in two variables can be used to graph the equation. However, the two solutions of an equation in two variables that are usually the easiest to find are those in which either the first or the second component is 0. The x-coordinate of the point where a straight line intersects the x-axis is given by is called the x-intercept of the line, and the y-coordinate of the point where a line intersects the y-axis is called the line-intercept. Using the intersection points to graph the equation is called the intersection graph method.

4. The slope of a line containing the points P1(X1, Of1) And p2(X2, Of2) is given by

Two lines are parallel if they have the same slope (m1= metro2).

Two lines are perpendicular if the product of their slopes is -l(m).1* M2= -1).

5. The point-slope form of a straight line with slope m that passes through the point (x1, Of1)Es

and and1-m(x-x1)

The slope-intercept form of a straight line with slope m and axis-intercept b is

y = meter x + second

6. A relationship governed by an equation of the form

y = kx (k is a constant)

is called direct variation.

7. A solution to an inequality in two variables is an ordered pair of numbers that, when substituted into the inequality, make the inequality a true statement. The graph of a linear inequality in two variables is a half plane. The symbols presented in this chapter are located inside the front cover.

(Video) SOLVING SYSTEMS OF EQUATIONS STEP-BY-STEP!

## Videos

1. Algebra Basics: Solving 2-Step Equations - Math Antics
(mathantics)
2. Algebra Basics: Solving Basic Equations Part 1 - Math Antics
(mathantics)
3. How to Solve One-Step Equations | One-Step Equation Steps | Math with Mr. J
(Math with Mr. J)
4. Graphing solutions to two-variable linear equations example 1 | Algebra I | Khan Academy
5. Problem Solving Straight Line Graphs - Mr Mathematics
(Jonathan Robinson)
6. Learn how to graph a quadratic
(Brian McLogan)
Top Articles
Latest Posts
Article information

Author: Kelle Weber

Last Updated: 22/06/2023

Views: 6420

Rating: 4.2 / 5 (53 voted)

Author information

Name: Kelle Weber

Birthday: 2000-08-05

Address: 6796 Juan Square, Markfort, MN 58988

Phone: +8215934114615

Job: Hospitality Director

Hobby: tabletop games, Foreign language learning, Leather crafting, Horseback riding, Swimming, Knapping, Handball

Introduction: My name is Kelle Weber, I am a magnificent, enchanting, fair, joyous, light, determined, joyous person who loves writing and wants to share my knowledge and understanding with you.