Ch. 4 Practice Test

Chapter Review 65Review Chapter 41. a. (~, YI)= (1,2), (x2,Y2)= (2,6)rise Y2- YI 6-2 4slope = -=-=-=-=4run X2-Xl 2-1 1b. (xl' YI)= ("5, '6), (x2, Y2) = ("8,2)2--6 8 -8slope=-=-= --8--5 -3 3c. (xl' YI)= ("2,3), (x2,Y2)= (1, 0)0-3 -3slope= 1_-2=3"=-1d. (xl' Y1)= (4, 'I), (x2,Y2)= (7,'1)-1- -1 0slope=-=-=O7-4 3e. (xl' Y1)= ("3,2), (x2,Y2)= ("7,4)4-2 2 -1slope=-=-= --7 - -3 -4 22. a. 0 inchesb. 2 incheslhourc.Time (hours)Amount of Snow onGround (inches)0 01 22 43 6d. The slope of the line is 2. The slope and rateof change (2 incheslhour) are the same. .y.o 2 3Time(hows)5 x

66 Chapter4: Systems of Linear Equations. 100f13. a. The rate 0 change 1S-=3 supp ements per-100dollar or - supplements per dollar3b. For every $3 decrease in price, Vitomax cansell 100 more supplements. Apply thisrate of change to the first ordered pair.$20 - $3 = $17, and 5000 + 100 = 5100 givinga second ordered pair of (17, 5100).c. The third ordered pair is:(17 - 3,5100 + 100) = (14,5200).d. The slope of the line is the same as the rate of-100change,m= -. 34. We could use the slope formula or notice that as xincreases by 2, Y increases by 3. This is a run of2and a rise of 3. Because slope is rise our slope =run5. Each equation is in slope-intercept form, so theslope is the coefficient of x and the constant b isthe initial value. The point (0, b) is where theinitial value appears on the graph, and we call thispoint the y-intercept.a. 4 The eqmitiony =3x + 5 has a slope of3 anda y-intercept of (0, 5). Graph (4) has apositive slope and appears to have ay-intercept of (0, 5).b. 3 The equation y = x has a slope of 1 and ay-intercept of (0, 0). Graph (3) has a positiveslope and appears to have a y-interceptof (0, 0).c. 1 The equationy ="x + 2 has a slope oCl anda y-intercept of (0, 2). Graph (1) has anegative slope and appears to have ay-intercept of (0, 2).d. 2 The equation y = "4 can be written asy = Ox- 4. The slope is 0 and the y-interceptis (0, -4). Graph (2) has a zero slope andappears to have a y-intercept of (0, "4).6. Plot the y-intercept (0, 3) and we can use the. rise 3slope to plot another pomt. Slope = - run = - 2 ;this gives run = 2 and rise = 3. Applying the runand rise to our starting location (0, 3), we get afinal location = (0 + 2, 3 + 3) = (2, 6).y- -7. a. 5inchesof snowb. 2 incheslhourc. Let x = the number of hours of snowfall andy =the accumulated inches of snow.From part (a) the initial value is b = 5 andfrom part (b) the slope is m = 2. Substitutingthese values into the slope-intercept formula, y= mx + b, we have y = 2x + 5 .8. a. The initial value is b = 1800 and the rate ofchange or slope is m = -300. Substituting thevalues for m and b into the slope-interceptformula, we have y = "300x+ 1800.b. We can quickly fmd ordered pairs that satisfythe equation from a TI-83 table. In theY= menu enter Yl = -300X + 1800. PressTBLSET and set TblStart = 0, ~Tbl= 1.Access the results by pressing TABLE. Theordered pairs are the same, so the equationchecks.@ Houghton Mifflin Company. All rights reserved.

Chapter Review 679. Substitute the values for m and b in the slopeinterceptfonnula.a. y =6x + 21b. Y =- x + 43c. y=2x+ld. y=-5x+.!.. 2e. y=x-3b. Y =mx + b when m=l, x=- 4, andy=55 =1 *C4)+b5=-4+b9=by = mx + b when m=1 and b=9y=x+9c. y=mx+b whenm=-3,x=l,andy=44=-3*1+b4 = -3 + b7=by = mx+ b whenm=-3 andb=7y= -3x+710. Put each equation in slope-intercept fonn. Thenthe slope is the coefficient of x and b is theconstant. The y-intercept = (0, b).a. y = 4x + 3; m = 4, y-intercept = (0, 3)d. y=mx+b-3=.!..*2+b2-3 = 1+ b-4=bwhen m=.!.. x=2 and y=-32' ,b. y= -x-I; m =-I,y-intercept= (0, -1)c. y = 3x - 5; m = 3, y-intercept'= (0, -5)d. y = x - 3; m = 1,y-intercept = (0, -3)-2 -2e. y= -x+4;m= -,y-intercept=(0,4)3 3r. y = -x + 2; m = -1,y-intercept= (0,2)11. 3x-4y=12+-3x +-3x-4y=-3x+12-4y = -3x+12-4 -4-3x 12y=-=4+-=43y=-x-3412. Substitute the values for m, x, and y into theslope-intercept fonnula, then solve for b. Writean equation by substituting the values for m and binto the slope-intercept fonnula.a. y=mx+b1=5*C2)+b1= -10+bl1=bwhen m=5, x=-2; andy=1Substitute for m, x and ySolve for by = mx+ b whenm=5and b=11y = 5x+ 11y=mx+b1y=-x-4 213. First, fmd the slope of the line connecting the twopoints with the slope fonnula, m = rise = Y2- Y. .run x2-~Then substitute the slope and either point into theslope-intercept fonnula and solve for b. Last,write an equation by substituting the values for mand b into the slope-intercept fonnula.6-4 2a m 2. -2-1-1-y =mx + b when m=2, x=l, and y=44 :: 2 *1+ b Substitute m, x, andy4=2+b2=by = mx + b when m=2 and b=2y=2x+23- -1 4b. m=-=-=45-4 1y = mx+ b whenm=4,x=5,andy=33=4*5+b3= 20+ b-17= by=mx+b whenm=4andb=-17y =4x-17@ Houghton MifflinCompany. All rights reserved.

68 Chapter 4: Systems of Linear Equations-8-6 -14 -7c. m=-=-= -6-2 4 2y=mx+b- 76= -*2+b26 = -7 + b13=by=mx+b-7y = -x+1327-4 3 -1d. m=-=-= --9-0 -9 3414. a. 1b.4yy=mx+b- 14= -*O+b34=0+b4=b-7whenm= - x=2 andY=62' ,-7whenm= -2andb=13- 1when m= - x=O andY =43' ,- 1y = mx + b when m= - and b=4 3- 1y= -x+43c. 3x+4y=12d. Substitute 1 for x into the equation3x+4y= 13.3 *1+ 4y = 12 Substitute fory3+4y =12 Solve fory4y=09y=- 4.. . 9The 0ther mUSICIanmust wnte songs fior- 41months or 2- months.4e. 3x+4y=12 wheny=O3x+4*0 =123x = 12x=4x-intercept = (4, 0)To complete the album, the first musicianmust write songs for 4 months when thesecond musician writes for 0 months.3x+4y=12 whenx=O3*0+4y=124y = 12y=3f.y-intercept = (0, 3)To complete the album, the second musicianmust write songs for 3 months when the firstmusician writes for 0 months.y.515. To find the x-intercept substitute 0 fory and solvefor x. To fmd the y-intercept substitute 0 for xand solve fory.a. x+ y=5x+0=5x=5x-intercept = (5, 0)x+y=50+y=5y=5y-intercept = (0, 5)- -yJ";. IDIOmoctK «(,5('\..r-;;:"'- - Ie t. )"\Ib. x-2y=4x-2*0=4x=4x-intercept= (4,0)x-2y=40-2y=4-2y = 4y= -2y-intercept= (0,-2)@ Houghton MifflinCompany. All rights reserved...

Chapter Review 69c.d.3x+2y=63x + 2 *0 = 63x=6x=2x-intercept = (2, 0)3x+2y=63*0+2y=62y=6y=3y-intercept = (0, 3)1-x-5y=1021-x-5*0=1021-x=102x=20x-intercept= (20,0)1-x-5y=1O21-*0-5y=1O2-5y=10y= -2y-intercept = (0, -2)1e. 2x--x=4312x - - *0= 432x=4x=2x-intercept= (2,0)12x--y = 4312*0--y=43- 1 -y=43y = -12y-intercept = (0, -12)16. a.f. 4x+5y=94x + 5 * 0 = 94x=99x=- 4x-intercept= (~, 0)4x+5y=94*0+5y=95y=99y=- 5y-intercept= (0, ~)-b. Intersection pointc. (1,3)17. Put each equation in slope-intercept fonn andenter the equations in the Y= menu. Choose awindow so that the graph shows the intersection.Press 2ndCALC, 5 for INTERSECT and pressENTER three times.Windows may vary: Xmin = '9.4, Xmax = 9.4,Ymin= '6.2, Ymax = 6.2Y1 = 3X + 1,Y2 = 2X + 2II't~li:r5.;:c~i(l1i~=1 f ~ IY='1Solution = (1, 4)y'","-i"Il"- t nI"'.1'-"-2 + - +4@ Houghton MifflinCompany. All rights reserved.

70 Chapter 4: Systems of Linear Equations18. YI = X- I, Y2= "2X-7 22. Let x = the first number and let y = the secondnumber.x+ y=78.2x - y = 29.3YI ="X + 78.2, Y2 =X-29.3Because the sum of the numbers is 78.2, we chose.......Il'IhrslQ:cti8)(a~= -2: / \1'1'= -3Solution = p, -3)19.a friendly first quadrant window wherethe maximum window variables exceed 78.2.Xmin = 0, Xmax = 94, Ymin = 0, Ymax = 100.Il'IhrslQ:cti8)(a~ = s:~.7 s: .::::..- 'I =21t. ..s: .........-The numbers are 53.75 and 24.45.No solution, this system is inconsistent becausethe lines are parallel. A comparison of the slopesshows both are the same, confirming parallellines.20. YI = "21X+ 14, Y2 = (l/2)X - 8The intersection is off the lower edge of thescreen; in the WINDOW menu changeYmin to "10.23. Let x =the number of lines of advertising and lety = the total cost of advertising (in dollars).YI = .65X + 7, Y2 = .52X + 8.30A small advertisement probably has less than 47lines and the cost is probably less than S50, so wechose a friendly first quadrant window ofXmin= 0, Xmax = 47, Ymin= 0, Ymax= 50.21.Il'IhrslQ:cti8)(a~=1.0232:S:S:B 1'1=-7. ..88372Solution"" (1.02, "7.49)--- ..Il'ItotrslQ:cti8)(a~=1(1. . . Y=13.~The newspapers charge the same amount for 10lines of advertising.24. y =-x+ 52x-y=6Substitute"x+ 5 fory into2x- y = 6.2x-Cx+5)=62x +x - 5=.6 Solveforx3x-5=63x = II11x=- 3Il'IhrslQ:cti8)(a~=3."'&"7Solution"" (3.67, 1.33)IY=1.3n3333Substitute!..!. for x iny = "x + 5, and solve fory.3-11y= -+53-11 15y= -+-3 34y=- 3Solution=( !..!.i 3' 3)@ Houghton Mifflin Company. All rights reserved.- - --

Chapter Review 7125. y=5x+lx+y=-l1Substitute5x+ 1fory intox +Y ='II.x+5x+l =-116x +1= -11 Solveforx6x = -12x= -2Substitute '2 for x iny =5x+ 1,andsolvefory.y=5*C2)+1y= -10+1y= -9Solution= p, -9)26. y = 3xy=7x-2Substitute3x fory intoy = 7x- 23x=7x- 2-4x =-2 Solve for x1x=- 227.Substitute .!..for x into y = 3x2..!..y=3 23y=- 2Solution = (~, %)3x + 4y = IS2x-4y = IS5x = 25x=5Add the equations3 * 5 + 4y = 7 Substitute 5 for x into 3x+4y=715 + 4y =7 Solve for y4y = -Sy= -2Solution = (5, -2)28. Multiply on both sides of the fIrst equation by '2so that the x-terms in the equations will beopposites of each other. Adding will theneliminate the x-term from the resulting equation.-2(x-5y) =-2(10) ~ -2x+ 10y =-202x+6y=4 ~ 2x+6y=416y = -16y= -12x+6*Cl)=42x+-6=42x = 10x=5Solution = (5, -1)Substitute -1 for y in 2x+6y=4Solve for x29. Multiply on both sides of the fIrst equation by '4and multiply on both sides of the second equationby 3. The x-terms are then opposites of eachother and adding eliminates the x-term from theresulting equation.-4(3x+7y)= -4(11) ~3(4x-5y)= 3C14) ~3x + 7 * 2 = 113x+ 14 = 113x = -3x= -1Solution= n, 2)-12x - 2Sy = -20l2x-15y= -42-43y=-S6y=2Substitute 2 fory in 3x+7y=11Solve for x30. We have an equation in slope-intercept form sowe chose to solve the system with substitution.y=9x-6x+2y= -12Substitute 9x - 6 for y into x + 2y = '12x+2(9x-6)= -12x + ISx -12 = -12 Solve for x19x-12=-1219x = 0x=Oy = 19· 6 - 6 Substitute 0 for x into y=9x - 6y= -6Solution = (0, -6)31. The equations are in standard form so we chose tosolve the system with elimination. Multiply onboth sides of the fIrst equation by 2 so that the y-terms will be opposites of each other. Addingthen eliminates the y-term from the resultingequation.2(2x+4y)=2C5) ~ 4x+Sy=-1O5x-Sy=-26 ~ 5x-Sy=-269x = -36x= -42 *C 4) + 4y = -5 Substitute - 4 for x in 2x+4y=-5-S + 4y = -5 Solve fory4y=33y=- 4Solution: = ( -4, ~)@ Houghton Mifflin Company. All rights reserved.

72 Chapter 4: Systems of Linear Equations32. a. x+ y=180b. y=2x-30c. We chose to solve this system withsubstitution.Substitute 2x - 30 for y into x + Y = 180x+2x-30=1803x-30 = 1803x = 210x=70Substitute 70 for x into y = 2x - 30y =2 * 70 - 30y =140-30Y =110The smaller angle is 70° and the larger angle is110°.33. This is a distance, rate, time problem. We solvedthis kind of problem before with one variable;now we use two'variables. This exercise issimilar to Example 1 in Section 4.4.The mouse has 2.25 seconds head start.Mouse TraveLs 2(%+ 2.25) metersCat TraveLs 5x metersLet x =the time (in seconds) that the cat runs;then x + 2.25 = the time that the mouse runs.The cat runs a distance of 5x meters. The mouseruns a distance of2(x + 2.25) m.Let y = the distance (in meters) traveled by eachanimal.Using the formula, distance = rate*time, we getthe system of equations:y=5xy = 2(x + 2.25)We chose to solve the system with substitution.Substitute 2(x + 2.25) for y into y=5x2(x + 2.25) = 5x2x + 2 *2.25 = 5x Solve for x2x + 4.5 = 5x4.5 = 3x1.5 = xThecat catchesthemousein 1.5= I.!.2seconds.Finish solving the system of equations, to checkthat the two animals run the same distance y.y =5 *1.5 Substitute 1.5 forx into y=5xy = 7.5 metersIf the solution checks, the mouse's distance willequal the cat's.Substitute 1.4 for x into y =2(x + 2.25)y =2(1.5+ 2.25)y=2*3.75y = 7.5 metersThe answer checks, both animals run 7.5 meters.34. a. ml = 5, m2= 5; the slopes are equal and they-intercepts are not equal, so the lines areparallel.b. ml = .!. , m2 = '4; the slopes are negative4reciprocals of each other so the lines areperpendicular.c. The equations in slope-intercept form are:y = '2x + 3, y = '2x + 3. ml = '2, m2= '2; theslopes and y-intercepts are equal, so theequations are the same.d. The equations in1slope-intercept9form are:Y =3x+l Y =-x--. 1, 5 5 ml = 3, m2= 5'; thelines are none of these.35. Find the slope of each line, then compare slopes.(xl' Y1)= (0, '3), (x2' Y2)= (2, 9)m,.= rise = Y2- Y1= 9 - -3 = 12 = 6run x2-Xl 2-0 2(xl' Yl) = (3, .5), (x2, Y2)= (12, 'I)-1-.5 -1.5 -1m=-=-= -2 12- 3 9 6ml = - -..!...; the slopes are negative reciprocals ofm2each other so the lines are perpendicular.Chapter 4 Test1. (xl' Yl)= ("6,1), (x2,Y2)= ("10,2)slope= rise= Y2- Yl =~ =J =- .!.run X2-XI -10--6 -4 42. a. This is a horizontal line on the x-axis.y- A@ Houghton Mifflin Company. All rights reserved.

Chapter Test 73b.- -y4. The slope is the rate of change, m = ~.2The y-intercept is (0, 250), therefore b = 250.Substituting the values for m and b into the slopeinterceptformula, y = mx + b, we have5y =-x+250.25. 2x+3y=6+-2x +-2xc.- -y/lA3y=-2x+6-2x+6y=-3-2 6y= -x+-3 3-2y= -x+23-2a. m= - 3d.yb. y-intercept = (0, 2)6. Substituting - ~ for m and 2 for b into the slope4-3interceptformula,we have: y = - x + 24-3. a. ~inches/year2b. For every 2 additional years, the tree heightincreases by 5 inches. Applying this to thefirst ordered pair (0, 250) we get a secondordered pair of (0 + 2, 250 + 5) = (2, 255).Applying the rate of change again gives a thirdordered pair of(2 + 2, 255 + 5) = (4, 260).Time (years)Height (inches)0 2507. y=mx+b-2=5*I+bwhenm=5,x=1,andy=-2-2=5+b-7 =by = mx+b whenm=5and b =-7y =5x- 78. 4x-5y=204- 5*0= 20wheny=O4x = 20x=5x-intercept= (5,0)4x-5y=20 whenx=O4*0-5y = 20-5y = 20y= -4y-intercept = (0, -4)y2 2554 260- -I..(0 )M@ Houghton Mifflin Company. All rights reserved.

74 Chapter 4: Systems of LinearEquations9. Let x = the number of miles, andlet y = the rental cost in dollars.The cost of renting the trucks is given by thesystem of equations:Yl = .08X + 19.95Y2 = .05X + 22.95Window may vary: Xmin = 0, Xmax = 188,Ymin = 0, Ymax = 100II'I~tr5';:C~il)l'I~=1CJ(I . . y=27.!l5____Both companies charge the same amount whentheir trucks are driven 100 miles.10. Yl =3X - 15/7Y2 = C6/7)X+ 3/7Window: ZOOM DecimalIl'Ihr5.;:c~il)l'I~=.66661i667 qy= - .1..28571Solution:::o (.67,-.14)11. y = 3x- ~ 76x + 7y =3Using substitution:Substituting 3x - ~ for y into 6x + 7y = 376X+7(3X-l;)=36x+ 2lx -15 = 3 Solveforx27x -15 = 327x =182x=- 3S b . . 2 .15u stitutlng- fiorx mto y =3x--3 7y=3*~-~ 3 715y=2--714 15y=--- 7 7- 1y= - 7. 2 - 1Solution = - -( 3' 7)12. 2x + 7y =- 53x + y =-17To solve by elimination multiply the secondequation by '7.2x+7y=-5-7(3x+y)=-7Cl7) ~2x+7y=-5-2lx-7y=119-19x = 114x= -63 *C 6) + y = -17 Substitute-6 for x in 3x+y=-17-18 + y = -17 Solve for yy=1Solution = ('6, 1)@ Houghton MifflinCompany. All rights reserved._.- --

Chapter Test 7513. Let x =the measure of the smaller angle, and let y= the measure of the larger angle.x+ y=1801-y=x-lO3To solve by substitution solve the secondequation for y and we gety =3x - 30.Substitute 3x - 30 for y into x + y = 180x + 3x - 30=1804x - 30=180 Solve for x4x=210x = 52.5Substituting 52.5 for x into y = 3x - 30Y = 3 *52.5 - 30Y = 157.5-30Y = 127.5The smaller angle is 52.5° and the larger angle is127.5°.14. The graph of a system of equations with nosolution form parallel lines.@ Houghton MifflinCompany. All rightsreserved.