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# Linear Equations in Two Variables

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary:

This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr.

This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. In this chapter, the emphasis is on the mechanics of equation solving, which clearly explains how to isolate a variable. The goal is to help the student feel more comfortable with solving applied problems. Ample opportunity is provided for the student to practice translating words to symbols, which is an important part of the "Five-Step Method" of solving applied problems (discussed in modules ((Reference)) and ((Reference))). Objectives of this module: be able to identify the solution of a linear equation in two variables, know that solutions to linear equations in two variables can be written as ordered pairs.

## Overview

• Solutions to Linear Equations in Two Variables
• Ordered Pairs as Solutions

## Solutions to Linear Equations in Two Variables

### Solution to an Equation in Two Variables

We have discovered that an equation is a mathematical way of expressing the relationship of equality between quantities. If the relationship is between two quantities, the equation will contain two variables. We say that an equation in two variables has a solution if an ordered pair of values can be found such that when these two values are substituted into the equation a true statement results. This is illustrated when we observe some solutions to the equation y=2x+5 y=2x+5 .

1. x=4,y=13;since13=2(4)+5istrue x=4,y=13;since13=2(4)+5istrue .
2. x=1,y=7;since7=2(1)+5istrue x=1,y=7;since7=2(1)+5istrue .
3. x=0,y=5;since5=2(0)+5istrue x=0,y=5;since5=2(0)+5istrue .
4. x=6,y=7;since-7=2(6)+5istrue x=6,y=7;since-7=2(6)+5istrue .

## Ordered Pairs as Solutions

It is important to keep in mind that a solution to a linear equation in two variables is an ordered pair of values, one value for each variable. A solution is not completely known until the values of both variables are specified.

### Independent and Dependent Variables

Recall that, in an equation, any variable whose value can be freely assigned is said to be an independent variable. Any variable whose value is determined once the other value or values have been assigned is said to be a dependent variable. If, in a linear equation, the independent variable is x x and the dependent variable is y y , and a solution to the equation is x=a x=a and y=b y=b , the solution is written as the

ORDERED PAIR      (a,b) (a,b)

### Ordered Pair

In an ordered pair, (a,b) (a,b) , the first component, a a , gives the value of the independent variable, and the second component, b b , gives the value of the dependent variable.

We can use ordered pairs to show some solutions to the equation y=6x7 y=6x7 .

### Example 1

(0,7) (0,7) .
If x=0 x=0 and y=7 y=7 , we get a true statement upon substitution and computataion.

y = 6x7 7 = 6(0)7 Isthiscorrect? 7 = 7 Yes,thisiscorrect. y = 6x7 7 = 6(0)7 Isthiscorrect? 7 = 7 Yes,thisiscorrect.

### Example 2

(8,41) (8,41) .
If x=8 x=8 and y=41 y=41 , we get a true statement upon substitution and computataion.

y = 6x7 41 = 6(8)7 Isthiscorrect? 41 = 487 Isthiscorrect? 41 = 41 Yes,thisiscorrect. y = 6x7 41 = 6(8)7 Isthiscorrect? 41 = 487 Isthiscorrect? 41 = 41 Yes,thisiscorrect.

### Example 3

(4,31) (4,31) .
If x=4 x=4 and y=31 y=31 , we get a true statement upon substitution and computataion.

y = 6x7 31 = 6(4)7 Isthiscorrect? 31 = 247 Isthiscorrect? 31 = 31 Yes,thisiscorrect. y = 6x7 31 = 6(4)7 Isthiscorrect? 31 = 247 Isthiscorrect? 31 = 31 Yes,thisiscorrect.

These are only three of the infintely many solutions to this equation.

## Sample Set A

Find a solution to each of the following linear equations in two variables and write the solution as an ordered pair.

### Example 4

y=3x6,ifx=1 y=3x6,ifx=1

Substitute 1 for x x , compute, and solve for y y .

y=3(1)6 =36 =3 y=3(1)6 =36 =3

Hence, one solution is (1,3) (1,3) .

### Example 5

y=154x,ifx=10 y=154x,ifx=10

Substitute 10 10 for x x , compute, and solve for y y .

y=154(10) =15+40 =55 y=154(10) =15+40 =55

Hence, one solution is (10,55) (10,55) .

### Example 6

b=9a+21,ifa=2 b=9a+21,ifa=2

Substitute 2 for a a , compute, and solve for b b .

b=9(2)+21 =18+21 =3 b=9(2)+21 =18+21 =3

Hence, one solution is (2,3) (2,3) .

### Example 7

5x2y=1,ifx=0 5x2y=1,ifx=0

Substitute 0 for x x , compute, and solve for y y .

5( 0 )2y = 1 02y = 1 2y = 1 y = 1 2 5( 0 )2y = 1 02y = 1 2y = 1 y = 1 2

Hence, one solution is ( 0, 1 2 ) ( 0, 1 2 ) .

## Practice Set A

Find a solution to each of the following linear equations in two variables and write the solution as an ordered pair.

### Exercise 1

y=7x20,ifx=3 y=7x20,ifx=3

(3,1) (3,1)

### Exercise 2

m=6n+1,ifn=2 m=6n+1,ifn=2

(2,11) (2,11)

### Exercise 3

b=3a7,ifa=0 b=3a7,ifa=0

(0,7) (0,7)

### Exercise 4

10x5y20=0,ifx=8 10x5y20=0,ifx=8

(8,20) (8,20)

### Exercise 5

3a+2b+6=0,ifa=1 3a+2b+6=0,ifa=1

#### Solution

( 1, 3 2 ) ( 1, 3 2 )

## Exercises

For the following problems, solve the linear equations in two variables.

### Exercise 6

y=8x+14,ifx=1 y=8x+14,ifx=1

#### Solution

( 1,22 ) ( 1,22 )

### Exercise 7

y=2x+1,ifx=0 y=2x+1,ifx=0

### Exercise 8

y=5x+6,ifx=4 y=5x+6,ifx=4

#### Solution

( 4,26 ) ( 4,26 )

### Exercise 9

x+y=7,ifx=8 x+y=7,ifx=8

### Exercise 10

3x+4y=0,ifx=3 3x+4y=0,ifx=3

#### Solution

( 3, 9 4 ) ( 3, 9 4 )

### Exercise 11

2x+y=1,ifx= 1 2 2x+y=1,ifx= 1 2

### Exercise 12

5x3y+1=0,ifx=6 5x3y+1=0,ifx=6

#### Solution

( 6, 29 3 ) ( 6, 29 3 )

### Exercise 13

4x4y=4,ify=7 4x4y=4,ify=7

### Exercise 14

2x+6y=1,ify=0 2x+6y=1,ify=0

#### Solution

( 1 2 ,0 ) ( 1 2 ,0 )

### Exercise 15

xy=0,ify= 14 3 xy=0,ify= 14 3

### Exercise 16

y=x,ifx=1 y=x,ifx=1

( 1,1 ) ( 1,1 )

### Exercise 17

x+y=0,ifx=0 x+y=0,ifx=0

### Exercise 18

y+ 3 4 =x,ifx= 9 4 y+ 3 4 =x,ifx= 9 4

#### Solution

( 9 4 , 3 2 ) ( 9 4 , 3 2 )

### Exercise 19

y+17=x,ifx=12 y+17=x,ifx=12

### Exercise 20

20y+14x=1,ifx=8 20y+14x=1,ifx=8

#### Solution

( 8, 111 20 ) ( 8, 111 20 )

### Exercise 21

3 5 y+ 1 4 x= 1 2 ,ifx=3 3 5 y+ 1 4 x= 1 2 ,ifx=3

### Exercise 22

1 5 x+y=9,ify=1 1 5 x+y=9,ify=1

#### Solution

( 40,1 ) ( 40,1 )

### Exercise 23

y+7x=0,ifx= y+7x=0,ifx=

### Exercise 24

2x+31y3=0,ifx=a 2x+31y3=0,ifx=a

#### Solution

( a, 32a 31 ) ( a, 32a 31 )

### Exercise 25

436x+189y=881,ifx=4231 436x+189y=881,ifx=4231

### Exercise 26

y=6(x7),ifx=2 y=6(x7),ifx=2

#### Solution

( 2,30 ) ( 2,30 )

### Exercise 27

y=2(4x+5),ifx=1 y=2(4x+5),ifx=1

### Exercise 28

5y=9(x3),ifx=2 5y=9(x3),ifx=2

#### Solution

( 2, 9 5 ) ( 2, 9 5 )

### Exercise 29

3y=4(4x+1),ifx=3 3y=4(4x+1),ifx=3

### Exercise 30

2y=3(2x5),ifx=6 2y=3(2x5),ifx=6

#### Solution

( 6, 21 2 ) ( 6, 21 2 )

### Exercise 31

8y=7(8x+2),ifx=0 8y=7(8x+2),ifx=0

### Exercise 32

b=4a12,ifa=7 b=4a12,ifa=7

#### Solution

( 7,40 ) ( 7,40 )

### Exercise 33

b=5a+21,ifa=9 b=5a+21,ifa=9

### Exercise 34

4b6=2a+1,ifa=0 4b6=2a+1,ifa=0

#### Solution

( 0, 7 4 ) ( 0, 7 4 )

### Exercise 35

5m+11=n+1,ifn=4 5m+11=n+1,ifn=4

### Exercise 36

3(t+2)=4(s9),ifs=1 3(t+2)=4(s9),ifs=1

#### Solution

( 1, 38 3 ) ( 1, 38 3 )

### Exercise 37

7(t6)=10(2s),ifs=5 7(t6)=10(2s),ifs=5

### Exercise 38

y=0x+5,ifx=1 y=0x+5,ifx=1

( 1,5 ) ( 1,5 )

### Exercise 39

2y=0x11,ifx=7 2y=0x11,ifx=7

### Exercise 40

y=0x+10,ifx=3 y=0x+10,ifx=3

#### Solution

( 3,10 ) ( 3,10 )

### Exercise 41

5y=0x1,ifx=0 5y=0x1,ifx=0

### Exercise 42

y=0(x1)+6,ifx=1 y=0(x1)+6,ifx=1

( 1,6 ) ( 1,6 )

### Exercise 43

y=0(3x+9)1,ifx=12 y=0(3x+9)1,ifx=12

### Calculator Problems

#### Exercise 44

An examination of the winning speeds in the Indianapolis 500 automobile race from 1961 to 1970 produces the equation y=1.93x+137.60 y=1.93x+137.60 , where x x is the number of years from 1960 and y y is the winning speed. Statistical methods were used to obtain the equation, and, for a given year, the equation gives only the approximate winning speed. Use the equation y=1.93x+137.60 y=1.93x+137.60 to find the approximate winning speed in

1. 1965
2. 1970
3. 1986
4. 1990

##### Solution

(a) Approximately 147 mph using ( 5,147.25 ) ( 5,147.25 )
(b) Approximately 157 mph using ( 10,156.9 ) ( 10,156.9 )
(c) Approximately 188 mph using ( 26,187.78 ) ( 26,187.78 )
(d) Approximately 196 mph using ( 30,195.5 ) ( 30,195.5 )

#### Exercise 45

In electricity theory, Ohm’s law relates electrical current to voltage by the equation y=0.00082x y=0.00082x , where x x is the voltage in volts and y y is the current in amperes. This equation was found by statistical methods and for a given voltage yields only an approximate value for the current. Use the equation y=0.00082x y=0.00082x to find the approximate current for a voltage of

1. 6 volts
2. 10 volts

#### Exercise 46

Statistical methods have been used to obtain a relationship between the actual and reported number of German submarines sunk each month by the U.S. Navy in World War II. The equation expressing the approximate number of actual sinkings, y y , for a given number of reported sinkings, x x , is y=1.04x+0.76 y=1.04x+0.76 . Find the approximate number of actual sinkings of German submarines if the reported number of sinkings is

1. 4
2. 9
3. 10

##### Solution

(a) Approximately 5 sinkings using ( 4,4.92 ) ( 4,4.92 )
(b) Approximately 10 sinkings using ( 9,10.12 ) ( 9,10.12 )
(c) Approximately 11 sinkings using ( 10,11.16 ) ( 10,11.16 )

#### Exercise 47

Statistical methods have been used to obtain a relationship between the heart weight (in milligrams) and the body weight (in milligrams) of 10-month-old diabetic offspring of crossbred male mice. The equation expressing the approximate body weight for a given heart weight is y=0.213x-4.44 y=0.213x-4.44 . Find the approximate body weight for a heart weight of

1. 210 mg
2. 245 mg

#### Exercise 48

Statistical methods have been used to produce the equation y=0.176x0.64 y=0.176x0.64 . This equation gives the approximate red blood cell count (in millions) of a dog’s blood, y y , for a given packed cell volume (in millimeters), x x . Find the approximate red blood cell count for a packed cell volume of

1. 40 mm
2. 42 mm

##### Solution

(a) Approximately 6.4 6.4 using ( 40,6.4 ) ( 40,6.4 )
(b) Approximately 4.752 4.752 using ( 42,7.752 ) ( 42,7.752 )

#### Exercise 49

An industrial machine can run at different speeds. The machine also produces defective items, and the number of defective items it produces appears to be related to the speed at which the machine is running. Statistical methods found that the equation y=0.73x0.86 y=0.73x0.86 is able to give the approximate number of defective items, y y , for a given machine speed, x x . Use this equation to find the approximate number of defective items for a machine speed of

1. 9
2. 12

#### Exercise 50

A computer company has found, using statistical techniques, that there is a relationship between the aptitude test scores of assembly line workers and their productivity. Using data accumulated over a period of time, the equation y=0.89x41.78 y=0.89x41.78 was derived. The x x represents an aptitude test score and y y the approximate corresponding number of items assembled per hour. Estimate the number of items produced by a worker with an aptitude score of

1. 80
2. 95

##### Solution

(a) Approximately 29 items using ( 80,29.42 ) ( 80,29.42 )
(b) Approximately 43 items using ( 95,42.77 ) ( 95,42.77 )

#### Exercise 51

Chemists, making use of statistical techniques, have been able to express the approximate weight of potassium bromide, W W , that will dissolve in 100 grams of water at T T degrees centigrade. The equation expressing this relationship is W=0.52T+54.2 W=0.52T+54.2 . Use this equation to predict the potassium bromide weight that will dissolve in 100 grams of water that is heated to a temperature of

#### Exercise 52

The marketing department at a large company has been able to express the relationship between the demand for a product and its price by using statistical techniques. The department found, by analyzing studies done in six different market areas, that the equation giving the approximate demand for a product (in thousands of units) for a particular price (in cents) is y=14.15x+257.11 y=14.15x+257.11 . Find the approximate number of units demanded when the price is

1. $0.12$0.12
2. $0.15$0.15

##### Solution

(a) Approximately 87 units using ( 12,87.31 ) ( 12,87.31 )
(b) Approximately 45 units using ( 15,44.86 ) ( 15,44.86 )

#### Exercise 53

The management of a speed-reading program claims that the approximate speed gain (in words per minute), G G , is related to the number of weeks spent in its program, W W , is given by the equation G=26.68W7.44 G=26.68W7.44 . Predict the approximate speed gain for a student who has spent

1. 3 weeks in the program
2. 10 weeks in the program

## Exercises for Review

### Exercise 54

((Reference)) Find the product. (4x1)(3x+5) (4x1)(3x+5) .

#### Solution

12 x 2 +17x5 12 x 2 +17x5

### Exercise 55

((Reference)) Find the product. (5x+2)(5x2) (5x+2)(5x2) .

### Exercise 56

((Reference)) Solve the equation 6[2(x4)+1]=3[2(x7)] 6[2(x4)+1]=3[2(x7)] .

x=0 x=0

### Exercise 57

((Reference)) Solve the inequality 3a(a5)a+10 3a(a5)a+10 .

### Exercise 58

((Reference)) Solve the compound inequality 1<4y+11<27 1<4y+11<27 .

3<y<4 3<y<4

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