58
Chapter 4: Systems of Linear Equations
Section 4.2
1. A linear equation in two variables may be written
in either slope-intercept form or standard form. If
the equation is solved for y in terms of x, it is in
slope-intercept form. If the equation has an x and
y term on one side of the equation and the
constant on the other side, than it is in standard
form.
y =8x -10
Slope-intercept form
+-8x + -8x
=-10 or 8x - y =10
-8x + y
Standard form
x+y=7
+-x +-x
b.
y
I
'
I
In
x
V"
II
y = -x + 7 Slope-interceptform
6x - 2y = 7
+-6x +-6x
c. On the graph, if we go from the x-intercept to
the y-intercept, we have a run = "2 and a
rise - 4
.
_
2
nse= 4 . SlOp e = -=-=
Standardform
run -2
-2y=-6x+7
-2y -6x 7
-=-+-2
5. a. x-intercept = (2, 0); the x-intercept is the
ordered pair where y is O.
y-intercept = (0, -4); the y-intercept is the
ordered pair where x is O.
-2
y = 3x- ~
2
d. The y-intercept is (0, "4) so b = "4. Substitute
2 for m and "4 for b in the slope-intercept
formula, y = mx + b; this gives y = 2x - 4.
-2
Slope-intercept
form
3. a. Because the equation is of the form
y = mx + b, it is in slope-intercept form.
b. m = "3;when an equation is in slope-intercept
form, the coefficient of x is the slope.
c. y-intercept = (0, 5); when an equation is in
slope-intercept form, the constant is the initial
value b and the y-intercept is the ordered pair
(0, b).
d. To find the x-intercept, if it exists, substitute 0
for y and solve for x.
y = -3x + 5
0= -3x + 5
3x=5
5
x=3
t
x-intercept= ( ' 0)
7. Substitute the values for m and b into the slopeintercept formula, y = mx + b.
3
a. y = mx + b when m = -5 and b = 4
3
y =-x + 4
5
b. y = mx + b when m = 2 and b = "1
y=2x-l
c. y=mx+bwhenm=Oandb=7
y = Ox+ 7
y=7
d. y=mx+bwhenm=
-1
-2 andb=O
-1
y= -x+O
2
-1
y= -x
2
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- ---
Section 4.2
9. We have two points, we are looking for the slope
and the value of b so that we can write an
equation for the line. Use a three step process.
(1) Use both points to find the slope using the
rise y-y
slope fonnula, = ~.
(2) Substitute the
run x2-~
slope and the values of x and y from either point
into the slope intercept fonnula, y = mx + b. Then
solve for b. (3) Write an equation by substituting
the values for m and b into the slope-intercept
fonnula.
a. (1) m=rise= Y2- YI =2-3 =.=.!.=-1
run
x2 -~
2-1
1
(2) y=mx+b
whenm=-I,x=l,andy=3
3 = -1 *1+ b Substitute for m, x, and y
3 = -1 + b
Solve for b
4=b
(3) y=mx+b
y = -Ix + 4
y=-x+4
6-1
b. (1) m=-=-=1
3 - -2
whenm=-1 andb=4
Substitute for m and b
5
5
(2) y = mx + b when m=l, x=3, and y=6
6=1*3+b
6=3+b
3=b
(3) y = mx + b when m=l and b=3
y=lx+3
y=x+3
-3--5
c. (1) m=-=-=
0-4
2
-1
2
-1
(2) Y = mx+ b whenm= -,2 x=O,andy=-3
-4
59
b. y will decrease by 22. This is the rate of
change or slope, m = '22.
c. Substitute the values for m and b into the
slope-intercept fonnula.
y = mx + b
when m = '22 and b = 3500
y = '22x + 3500
13. a. From the table the low temperature is 66°F.
Temperature
coF)
Hour of Day
Elapsed
hours since
2 P.M.
10:00 P.M.
8
75
Il:OO P.M.
9
73.5
12.00 A.M.
10
72
1:00 A.M.
Il
70.5
2:00 A.M.
12
69
3:00 A.M.
13
67.5
4:00 A.M.
14
66
b. We have the rate of change m and a point
(x, y). Substitute the values for m, x, and y
into the slope-intercept fonnula and solve for
b. Then use the values for m and b to write an
equation.
Let x = the number of hours since 2 P.M. and
let y = the temperature in OF.
y=mx+b
whenm=-1.5,x=8,andy=75
75 = -1.5*8+b
75=-12+b
87 =b
y=mx+b
whenm=-1.5 andb=87
y= -1.5x+87
This equation is valid for elapsed times
between 0 and 14 hours.
c. Windows may vary: Xmin = 0, Xmax = 18.8,
Xscl = 2, Ymin= 65, Ymax= 90, Yscl= 4
1=-1.5:3+87
(3) y = mx + b
-1
y= -x-3
2
4-4
0
d. (1) m=-=-=O
-1- 3 -4
(2) y = mx+ b whenm=O,x=3,andy=4
4=0*3+b
4=b
(3) y = mx+ b whenm=Oandb=4
y = Ox+ 4
y=4
11. a. y = 3500 feet; this is the parachutist's initial
height, b = 3500.
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3=1... . . \'=&&
~.
.
4 A.M. is 14 hours after 2 P.M, so trace the
graph to x = 14 to fmd the low temperature of
66°F.
d. y= -1.5x+87 whenx=14
y= -1.5*14+87
y= -21+87
y=66
The exactlowtemperatureis 66°F.
60
Chapter 4: Systems of Linear Equations
Skills and Review 4.2
15. a. Work = rate * time; Paul complt,tes ~ job
in 1 hour and .!.x jobs in x hours.
3
b. Jo:m completes .!.job in 1 hour and .!.Y jobs
2
2
17. a. (XhYI) = (1, 2), (x2,Y2)= (5, 7)
run = X2 - XI = 5 - 1 = 4,
rise = Y2 - YI = 7 - 2 = 5
b. slope = rise = Y2 run
x2-~
Y1
=~
4
in Y hours.
c. distance = ../run2+ rise2
c. Paul's jobs + Joan's jobs = 6
1
1
-x+-y=6
3
2
1
d. -Y=
2
-1
-x+6
3
-2
Y = -x+12
3
="/42 +52
= v'16+ 25
=J4i
",,6.4
-1
Add - onbothsides
3
Multiply by 2 on both sides
e. Friendly windows may vary: Xmin = 0,
Xmax = 18.8,Xsc1 = 2, Ymin= 0, Ymax = 15,
Ysc1= 2.
Y1= C2/3)X + 12
y-intercept
./
(0. 12)
1
1
19. 1--(x+9)=4(5--x)
3
4
1
1--x-3 = 20-x
3
-2-.!.x=20-x
3
+x +x
-2+3.x = 20
3
+2 +2
2
f. To find the x-intercept substitute 0 for Y and
solve for x.
1
1
-x+-y=6
3
2
1
1
-x+-*0=6
3
2
1
-x=6
3
x=18
The x-intercept is (18, 0); it takes Paul 18
hours to complete 6 jobs alone.
To find the y-intercept, substitute 0 for x and
solve for y.
1
1
-x+-y=6
3
2
1
1
-*0+-y=6
3
2
1
-y=6
2
Y =12
The y-intercept is (0, 12); it takes Joan 12
hours to complete 6 jobs alone.
g. The slope is - 3..
3 For each 3-hour increase in
Paul's hours, Joan's hours decrease by 2. Or,
for each 3-hour decrease in Paul's hours,
Joan's hours increase by 2.
-x = 22
3
3 2
3
-*-x=-*22
2 3
2
x=33
21. a. The length of the diagonal is the length of the
hypotenuse c of a right triangle whose other
sides measure a = 6 m and b = 9 m. Use the
formula c = .,/a2 + b2
c = "/62 + 92
= v'36 + 81
=.Jill
""
10.8
Diagonal"" 10.8 m
.Jill m
b. Diagonal=
23. (4*83)-(4*3)
= 4 * (83 - 3)
= 4 *80
.
=320
25. -24- -8-6+ 22
= -24+8-6+22
= -16-6+22
=-22+22
=0
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