Download Section 5.3

80
Chapter 5: More Applications of Linear Equations
d. Solving by elimination:
-4(x+y)=-4(3)
~ -4x-4y=-12
4x+6y=13
4x+6y=13
~
2y=1
Section 5.3
1. a. Weare looking for the cost per pound of the
mixture. We can write this as
cost
y=-
pound
$28
1
x+-=3
2
10 pounds
= $2.80
1
2
1
x=2-=2.5
2
per pound
b. Peanuts
c. Peanuts, because $2.80 is closer to the $2
peanut price than it is to the $5 cashew price.
e. There is 2.!. pounds of chocolate raspberry
2
and .!. pound of French vanilla.
2
9. a.
=
+
..mJ
yml
b. x + y '" 30
c. .15x + .40y
xm!
=.25(30)
d. (1) Solve eaeh equation for y in tenns of x
(2) Enter the equations in the Y= menu on the
calculator.
(3) Find a window that includes the
intersection of the two lines.
(4) Use TRACE or 2ndCALC 5: intersect,
to fmd the intersection.
600mI
ym!
b. Let x ='the amount (in mL) of 10% solution
and lety" the amount (in mL) of 40%
solution.
c.
x + y = 600
Solutionequation
.10x+ .40y =.31(600) Alcoholequation
d. YI = -x + 600,Y2 = -.25X+ 465
Windowsmay vary:Xmin= 0,Xmax= 940,
Ymin= 0, Ymax= 700
S. Distance equation 4x + lOy = 16: 4x is the
distance Jason travels, 10y is the distance Nathan
travels, and 16 is their combined distance.
Time Equation x
=y + .!.
2 :x is the number of.
hours Jason travels; y is the number of hours
1
Nathan travels, and - hour represents Nathan's
2
later start time.
7. a. Let x = the amount of chocolate raspberry (in
pounds) andy = the amount of French vanilla
(in pounds).
b.
c.
Chocolate
Raspberry
French
Vanilia
Rate
($/pound)
4
6
Weight
(pounds)
x
y
3
Cost ($)
4x
6y
13
Mixture
II'I~.rcr5IQ:C~iO)fI
H=lIIO
.
x= 180,y=420
.
'1'="20
e. We need 180 mL of the 10% solution and 420
mL of the 40% solution.
f. Quick check: There should be quite a bit more
of the 40% solution in the mixture because
31% is much closer to 40% than it is to 10%.
Thorough check: The numbers should meet
the conditions of the problem.
x + y = 6 Weight equation
4x + 6y =13 Moneyequation
@ Houghton Mifflin Company. All rights reserved.
--
- -
Section
11. a.
Total driving tlme 6 hr 30 mln
WIchita
Total distance
..t
St Louis
,I
446 mI
I.~
61xmlles
,,~
72ymlles
b. Let x = the time in hours Thelma drives and let
y
= the time
x+y
61x + 72y
Substitute 5600
in hours Louise drives.
-
- - 22
x=2
y = 6.5- 2 Substitute:2 for.x intoy=6.5- x
y-4.5
y
4
12
Rate
($/ticket)
solute
x
y
Cost ($)
4x
12y
3040
x=18
We need 18 g of hydrochloric acid. The
remainder of the 300 g of solution is water. 300 g
- 18 g = 282 g. We need to mix 282 g of water
with 18 g of hydrochloric acid to produce 300 g
of solution.
= 3040
Ticket equation
Money equation
Solving the equations for y in terms of x, we have
Y1 = 'X + 600, Y2 = Cl/3)X + 760/3
Windows may vary: Xmin = 0, Xmax = 940,
Ymin = 0, Ymax = 700
II'J~tr5.;:c~i(Jr.
~=S2:(J
x = 520, y = 80
The company sells 520 ordinary seats and 80
special seats.
@ Houghton Mifflin Company. All rights reserved.
-24
7
7x = 3.2 * -24
7x
Let x = the number of ordinary seats and
let y = the number of special seats.
= 600
= percent. solution
x = .06 '"300
19. 2-=
3.2
Number
of tickets
Substitute 4100 for x
into y=5600 - x
17. Find the part of the solution that is hydrochloric
acid (solute).
part = percent*wholc
Mixture
600
x+y
4x + 12y
= 5600 - 4100
Solve fou
Skills and Review 5.3
13.
Special
Seats
in the second equation.
y = 1500
The couple borrows $4100 from the 7.5% source
and $1500 from the 11.25% source.
d. Thelma drives for 2 hours and Louise drives
for 4.5 hours.
Ordinary
Seats
- x for y
y =5600-x
.075x + .1125(5600 - x) = 476.25
.075x+ 630-.1l25x = 476.25
-.0375x + 630 = 476.25
-.0375x = -153.75
x = 4100
Time equation
= 446 Distance equation
Ilx
81
15. Let x = the amount of money (in dollars)
borrowed at the lower rate, and let y = the amount
of money (in dollars) borrowed at the higher rate.
x + y = 5600 Money equation
.075x + .1125y = 476.25 Interest equation
Solving by substitution:
Solve the first equation for y in terms of x
=6.5
c. Solving by substitution:
y = 6.5 - x Solve the first
equation fory
in terms of x
Substitute
61x + 72y(6.5 - x) = 446
6.5-x fory
in the second
equation
Solve for x
61x + 468 -72x = 446
12x.. 46S 446
5.3
x
21.
=-76.8
Use the cross multiplication
property
Solve for x
= -76.8 '" -10.97
7
9x-4y=36
9x - 4 * 0 = 36 Substitute 0 for y
9x = 36
x=4
x-intercept = (4, 0)
9x-4y=36
9. 0 - 4y = 36 Substitute 0 for x
- 4y = 36
y= -9
y-intercept = (0, '9)
82
Chapter 5: More Applications of Linear
23. 4 - (x + 3) =2(5x -1)
4-x-3 =10x-2
1- x = lOx = 2
1=l1x-2
3 = 11x
3
x=
27
11
25. The missing side of this right triangle is the
hypotenuse c. A formula for finding the length of
the hypotenuse is:
c =.Ja2 + b2
=.J62 + 82
=../36+ 64
=J100
=10 inches
-