ChapterReview 49

d. 1.2-x=.7-.5x+.5x +.5x

1.2-.5x =.7

+-1.2 +-1.2

- .5x= -.5

- .5x - .5----.5 -.5

x=1

e. 6=2x+9

+-9 +-9

-3 = 2x

-3 2x2 2-3- = x orx = -1.52

c. 5x+ 7= -23

+-7 +-7

5x = -30

5x -30-=-5 5x= -6

f. 7x-8=5x+13+ - 5x +- 5x

2x-8=13

+8 +8

2x = 21

2x 21

2 221

x=-=10.52

1 7g. -x+6=-x-212 2

- 1 - 1+ -x + -x~~

6 =3x-21

+21 +21

27 =3x

27 3x

3 39=x

Review Chapter 3

1. a. 3x+8=11

+ -8 +-8

3x=33x 3-=-3 3x=1

b. 7+x=3-4x+4x +4x

7+5x=3+-7 +-7

5x= -45x -4-=-5 5

-4x- ---8

- 5 - .

2 1h. -x-l=6+-x

3 3- 1 - 1

+ -x + -x---L ---L1-x-l=63

+1 +11-x=731

3*-x=3*73

x=21

@Houghton MifflinCompany. Allrights reserved.

50 Chapter 3: Linear Equations

2. 2x- 3= x + 6 whenx = 9

2*9- 3 = 9 + 6 Substitute 9 for x

18 - 3 = 1515=15

Simplify on both sidesThe solution x = 9 checks

3. a. Time(hours)1234t

Distance (miles)50

10015020050t

b. Use the formula, distance = rate*time; the rateis 50 mi1es/hourand the time is t hours.Therefore the distance is 50t miles.

c. The solution represents the number of hoursthat it takes to travel a distance of 120 miles ata rate of 50 miles/hour.

d. 120= 5t120 50t-=-50 5012-=t5

2.4= tIt takes 2.4 hours or 2 hours 24 minutes totravel a distance of 120 miles, moving at a rateof 50 mi1es/hour.

4. a. K = C + 273whenK = 373373= C+ 273

b. 373= C+ 273

+-273 +-273100= C

Waterboilsat 100°C.

5. T = 38 + 1; find 8 when T = 28.28 = 38 + 1

+ -1 + -1

27 = 38

27 38---3 39=8

A row made of 28 toothpicks has 9 squares.

6. a. The cost of the other club is the sum of thefIxed cost and variable cost; 2.50 + 1.50d.

b. 4 + 1d= 2.50 + 1.50d

c. 4 + 1d =2.50 + 1.50d

+-ld + -ld

4 = 2.50 + .50d

+ -2.50 +-2.50

1.50 =.50d

1.50 .50d---.50 .50

3=dThe feesat eachclubare thesamewhenexactly3 CDsarepurchased.

7. a. 3x+2-x-5=2x-3

b. 2(x + 1)+ 3x - 4= 2x + 2 + 3x - 4=5x- 2

c. x-(6-4x)+2x=x+ -1*(6-4x)+2x=x-6+4x+ 2x=7x-6

d. 3(2x-4)+x2 -12=6x-12+x2-12=x2+6x-24

e. 3y_x2 - y+4x2=3x2 + 2y

1 1f. -+x+-+2x

3 61 1

=x+2x+-+-3 6

2 13x+- +-

6 63=3x+-61=3x +-2

1 1g. -x--x2 3

3 2=-x--x

6 61=-x6

3 1h. -(2x+8)--x

4 23 1=-x+6--x2 22=-x+62

=x+6

@ Houghton Mifflin Company. All rights reserved.

Chapter Review 51

= 3x + -3 + Ix + 3=4x

8.3x-(3-x)+3= 3x + -1*(3 -x) + 3 Using the special property

of -I, c lets us change thesubtraction in front of the leftparenthesis to addition if weinsert a -1

b. 5(x - 2) + 9 = 7(x + 3)5x-l0 + 9= 7x+ 21

5x-l=7x+21+-5x +-5x

'I =2x+21+-21 +-21

-22 = 2x-22 2x---2 2'11 =x

LS,RSLS

g. 2 + 3(4x+ 6) =x+ 7 -(2x- 8)2 + 12x + 18 =x + 7 - 2x + 8

12x+20= -lx+1513x+ 20 = 15

13x= -5- 5x= -13

A

M

LS,RS

2 1h. -(x-9)+-x=7x-15

3 32 1-x-6+-x=7x-153 3

x-6=7x-15-6=6x-159=6x3- = x or x = 1.52

A

10. a. x+8+x+8=262x+16=26

2x = 10x=5

d. 4(x -5)+ x = 10+ 6x-704x-20+x = 6x= 60

5x- 20= 6x-60-lx-20= -60

-lx=-40x=40

11. P = 2L + 2W ; fmd L when P = 20 inches andW = 4 inches.20 = 2L + 2 *420 = 2L + 812= 2L6=L

The rectangle has a length of 6 inches.

12. A = 4(x+3); find x when A = 15 cm2

15=4(x+3)15= 4x+ 123=4x3-=x4

b. 4x - 6+ 6x = 5+3x-llOx- 3= 4+3x

7x=7x=1

c. x+(x+30)+(2x+l0)=180x+ x+30+ 2x+ 10= 180

4x+40=1804x = 140x=35

e. 1- (x +5)= 3(x - 2)+ 4x1+-1(x+5)= 3x-6+4x1+ -Ix + -5=3x - 6+4x

-lx-4=7x-6-8x-4=-6

-8x = -21x =- =.254

@ Houghton MifflinCompany. All rights reserved.

9. a. 2x+ 8+ x = 323x + 8= 32 RS

+-8 +-83x= 24 A3x 24 M---3 3x=8 LS,RS

52 Chapter 3: Linear Equations

13. a. Less than 1 hour; in 1 hour the combineddistance of the two trains is 45 + 55 or100 miles.

b. The northbound train travels a distance of 45tmiles.

b.

c. 45t + 55t =36

d. 45t + 55t =36lOOt=36

9t=-=.3625

Convert hours to minutes

36 t L 60 minutes 21 6'

. J14'Ul* 1)16tif - . mmutesRound to the nearest minute and find the timeof day. 2:00 P.M. + 22 minutes = 2:22P.M.Atabout 2:22 P.M., the trains are 36 miles apart.

14. a. x + 20 represents the measure of angle B and x+ 70 the measure of angle C.

b. x+x+20+x+ 70=180

c. Let dl'= the distance (in blocks) from (2,3) to(5,1).Let d2= the distance (in blocks) from (5,1) to(1, -2).

c. Both unknown angles measure 45°,

d. distance = -/run2 + rise2

(Xl>YI) (X2,Y2)(2,3) ("5, 1)run=X2 -Xl = '5 - 2 = '7,rise = Y2 - YI = 1 - 3 = '2

d1=~C7)2 +c2i

(XI>YI) (X2,Y2)("5, 1) (1, '2)run = X2- XI = 1 - -5 = 6,rise = Y2- YI = '2 - 1 = -3

d2 = ~62 + C3)2

e. d1= ~C7)2 + c2i

= --149+4

=-./53",,7.3

d2 = ~62 + C3)2

= --136+9

=J45",,6.7

c. x+x+20+x+70=1803x+90= 180

3x = 90x=30

d. The measure of angle A is 30° angle B is 30+ 20 = 50°,angleCis 30+ 70 = 100°,

15. a. X represents the measure of angle B and angleC90+x+x= 180

b. 90 + X + X = 18090 + 2x = 180

2x=90x=45

16. a. f. We need the sum of the distances dl and d2.The total distance is

=d1 + d2",,7.3+6.7=14The pilot flew approximately 14 blocks.

g. The answer checks with an approximate countof the blocks in the figure from part (b).

b. Let X=the measure of each angle in anequilateral triangle. Then we havex+x+x =180

3x =180x=60

Each angle of an equilateral triangle measures60°.

17. a. Find the sum of the distances between thepoints (2,3) and (5,1) and between r5,1) and(1,-2).

@Houghton Mifflin tompany. All rights reserved.

y

(2 3)

- A

1-

Chapter Review 53

18. (1) Find the dimensions of a tennis court withperimeter 228 feet and a length that is 6 feetmore than twice the width.

20. a.

(1) Find the time it takes Jones to catch heropponent.

(2)Jones Travels 8.8t mlles

2x+ 6 Opponent Travels 8.6(t +.3) mlles

(3) Let x = width, then 2x + 6 = length.

(4) Perimeter = the sum of the lengths of the sidesof the rectangle.228 =x + 2x + 6 + x + 2x + 6

(3) Let t =the time (in seconds) that Jones runs;then t + .3 = the time that Jones' opponentruns.

(6) The width is 36 feet and the length is2*36 + 6 or 78 feet.

(7) The sum of the lengths of the sides should be228 feet.36 feet + 78 feet + 36 feet +78 feet = 228 feet,the answer checks.

(4) When Jones catches her opponent they haverun the same distance. In t seconds at a rate of8.8 mis, Jones runs a distance of 8.8t m.In t + .3 seconds at a rate of 8.6 mis, Jones'

opponent runs a distance of 8.6(t + .3) m.Set their distances equal to each other.8.8t =8.6(t + .3)

(5) 8.8t =8.6(t + .3)8.8t =8.6t + 2.58

.2t =2.581=12.9

(5) 228 =x + 2x + 6 + x + 2x + 6228 =6x + 12216 =6x

36=x

(3) Let t = the number of minutes Sue uses herphone each month.

(4) At 6 cents per minute for t minutes of calls,the variable cost is .06t. The fixed cost is$11.30 and the total cost is $40.total cost =variable cost + fixed cost.40 =.06t + 11.30

(6) Jones runs for 12.9 seconds to catch heropponent.

(7) If the time is correct, Jones and her opponenthave run the same distance. Jones runs for12.9 seconds at a rate of 8.8 mis, for a distanceof 113.52 m. Her opponent runs for 12.9 + .3or 13.2 seconds at a rate of 8.6 mis, for adistance of 113.52 m. Their distances are thesame, so the answer checks.

19. (1)Find how much time Sue spend on the phoneeach month for a total of $40.

(2) The total cost is the sum of a variable cost anda fixed cost.

(5) 40 =.061+ 11.3028.70 =.06t

478.33""t

b. distance = rate*time= 8.8*12.9= 113.52 mJones catches her opponent at a distance of113.52 m.

c. The race is 200 m so Jones wins the race.

(7) At $,.06per minute, 478 minutes costs $28.68.Add the fixed cost of $11.30 for a total of$39.98. This is very close to $40, so theanswer checks.

21. a. Arithmetic; add 4 to each term to find the nextterm: 22, 26, 30.

b. Geometric; multiply each term by 3 to fmd thenext term: 405, 1215,3645.

(6) Sue can use the phone for approximately 478minutes each month.

c. Odd terms increase by 5 and even terms are 0:0,17, O.

d. Arithmetic; subtract 2 from each term to fmdthe next term: 11,9,7.

@ Houghton Mifflin Company. All rights reserved.

54 Chapter 3: Linear Equations

22. a. a. = a1+ (n -l)d ; find 010when a1= 1,d= 7,alO means n = 10.alO= 1 + (10 - 1)*7

010= 1 + 9*7alO = 1 + 63alO = 64

5.9

F = -C +32; findC whenF = 90°59

90=-C+3259

58=-C5

290 = C9

32.2"'CTheCelsiustemperatureis approximately32.2°whentheFahrenheittemperatureis 90°.

b. 0. =a1+(n-1)d; fmdd when a1= -3,a9 = 5, Ii9 means n = 9.5 = '3 + (9-l)d5 = -3+ 8d8=8dl=d

c. a. = a1+ (n -l)d ; find a1when d = 11,a14 = 151,al4 means n = 14.151= a1+ (14 -1)*11151 = al + 13*11151 =a1 + 1438=a1

6. a. 20-~13

Chapter 3 Test

1. 7x+5=2x-57x+5=-5

5x = -10x= -2

b. 20- ~ whenx = 5 miles13

5= 20--

13",19.6Approximately19.6gallons.

c. At 13miles/gallon,20gallonsof gaswillallowtheSUVto go 260miles.

7. a. Let / = thenumberof hoursit tookfortheboats to meet.

2. 6-(~X+2))=9+~X-34 2

6--x-2=9+-x-35 5

4 24--x=6+-x

5 56

4=6+-x5

-2=~x5

b. This is a distance, rate, time problem. Wehave the total distance the boats traveled andthe rate of each boat. After finding the time ittook for the boats to meet, we can fmd thetime of day.In t hours, at a rate of 15.5 mi1es/hour, theCarpathia traveled a distance of 15.5t miles.In t hours, at rate of 1.5 mi1es/hour, the Titanicdrifted a distance of 1.5t miles. The sum oftheir distances is 58 miles.15.5/ + 1.5t = 58

-10-=X6

-5-=x3

c. 15.5t+1.5t= 5817t= 58

58t=-",3.41

17It took the boats about 3.41 hours or 3 hours25 minutes to meet. Add this elapsed time tothe start time of 12:25 A.M; the boats met atabout 3:50 A.M.

33. 6x - 1 = 2x + 5 when x = -2

3 ' 36*--1=2*-+5

2 2,9-1=3+5

8 = 8 Thesolutionx = ~ checks2

4. 2x+9 -5x =4(x -3)-3x+9 =4x-12

9=7x-1221= 7x3=x

@ Houghton MifflinCompany. All rights reserved.

8. a. c

A B .

Let x represent the measure of angle A, then2x represents the angle B and angle C.The sum of the angles is 180°.x + 2x + 2x = 180

b. 5x = 180x=36

c. Angle A measures 36°, and angles B and Cmeasure 2*36 = 72°.

9. a.Doug Travels 91 miles

Lance Travels 7.8(1 +.25) miles

Let t = the time(inhours)thatDougruns,thent + .25 = the time that Lance runs (15 minutesis one-quarter hour or .25 hours).When Doug catches Lance they have run thesame distance.In t hours, at a rate of 9 miles/hour, Doug runsa distance of 9t miles.In t + .25 hours, at a rate of7.8 miles/hour,Lance runs a distance of7.8(t + .25) miles.Set their distances equal to each other.

9t =7.8(t + .25)9t=7.8t+1.95

1.2t = 1.95t = 1.625

In 1.625 hours or 1 hour 37 minutes 30seconds, Doug catches Lance.

b. Yes. Lance runs about 8 miles per hour. In 15minutes, or one-quarter of an hour, Lance getsabout a 2-mile head start on Doug.Doug is faster than Lance by a little more than1 mile/hour. Therefore, in less than 2 hoursDoug will have made up Lance's 2-mile headstart.

@ Houghton Mifflin Company. All rights reserved.

Chapter Test 55

10.

3x+4Let x = the length (in cm) of a leg of the isoscelestriangle, then the remaining side is 3x + 4.The 20 cm perimeter is the sum of the length ofthe sides.20 =x+x+3x+4

20 =5x+4

16 =5x

16-=x5

3.2=xThelengthof eachleg is 3.2cmandtheremainingsideis 3*3.2+ 4 or 13.6cm.