Monday’s Warm Up
Solve for x :
24  x  7
x  7  24
7 7
x  17
Objective
By the end of today’s lesson, you will be able to
solve an equation for a particular letter, given
that the equation contains multiple letters
Example #1
Solve for x
Solve for x
24  x  7
x  7  24
d  x p
x p d
x  17
xdp
Example #2
Solve for t
Solve for t
35  5t
d  rt
rt  d
5t  35
t7
d
t
r
Example #3
Solve for l
Solve for l
60  l  5  4
V  lwh
5  4  l  60
lwh  V
20l  60
l 3
V
l
wh
Example #4
Solve for b
Solve for b
10  b  3  4
z  br  p
z  br  p
br  p  z
br p  z

r
r
pz
b
r
10  3b  4
3b  6
b  2
Example #5
Solve for t
Solve for t
2 5  t  35
10  t  35
t  25
t  25
2s  t  r
2s  t  r
t  r  2s
1  t   1  r  2s 
t  r  2s
Example #6
Solve for p
p
45 
6
p
 45
6
6 p
  45  6
1 6
p  270
Solve for p
p
h
n
p
h
n
n p
  hn
1 n
p  hn
Example #7
Solve for r
Solve for r
78.5  2 3.14 r
C  2 r
6.28r  78.5
2 r  C
2 r C

2
2
C
r
2
6.28r 78.5

6.28 6.28
r  12.5
Example #8
Solve for l
Solve for l
160  2l  2 15
P  2l  2w
2l  30  160
2l  2 w  P
2l  130
2l  P  2w
l  65
P  2w P
l
 w
2
2
Example #9
Solve for x
Solve for x
5 x  4  3x  32
ax  b  cx  d
ax  cx  b  d
5 x  3 x  4  32
2 x  4  32
2 x  28
x  14
 a  c x  b  d
 a  c x  d  b
a  c x  d  b
a  c a  c
Example #9
Solve for x
Solve for x
5 x  4  3x  32
a  c x  d  b
a  c a  c
5 x  3 x  4  32
2 x  4  32
2 x  28
x  14
d b
x
ac
Example #10
Solve for b1
Solve for b1
120  5b1  3
1
2  5 b1  3  120
A  12 h b1  b2 
1
2 h  b1  b2   A
1
2
2
1

1
2
5b1  3  120  2
5b1 3
240
 5
5
b1  3  48
b1  45
2
1
 h b1  b2   A  2
1
2
h b1  b2 
h

2A
h
b1  b2  2hA
b1 
2A
h
 b2
Motion Problem Set Ups
3. Lois rode her bike to visit a friend. She traveled at 10 mi/h.
While she was there, it began to rain. Her friend drove her
home in a car traveling at 25 mi/h. Lois took 1.5 h longer to go
to her friend’s than to return home. How many hours did it
take Lois to ride to her friend’s house?
Rate Time Distance
to by bike 10mph t  1.5 10  t  1.5
home by car 25mph
t
25  t 
10 t 1.5  25 t 
Motion Problem Set Ups
4. Fred begins walking toward John’s house at 3 mi/h. John
leaves his house at the same time and walks toward Fred’s
house on the same path at a rate of 2 mi/h. How long will it be,
in minutes, before they meet if the distance between the
houses is 4 miles?
Rate Time Distance
Fred 3mph
t
3t 
John 2mph
t
2 t 
3 t   2  t   4
Motion Problem Set Ups
5. May rides her bike the same distance that Leah walks. May
rides her bike 10 km/h faster than Leah walks. If it takes May
1 h and Leah 3 h to travel that distance, how fast does each
travel?
Rate Time Distance
May r  10 1hr
 r  10 
Leah
r
3hr
3 r 
 r 10  3 r 
Motion Problem Set Ups
6. A train leaves the station at 6:00 P.M. traveling west at 80
mi/h. On a parallel track, a second train leaves the station 3
hours later traveling west at 100 mi/h. At what time will the
second train catch up with the first?
Rate Time Distance
First Train
80mph
t
80t
Second Train 100mph t  3 100  t  3
80 t   100  t  3
Motion Problem Set Ups
7. At 10:00 A.M., a car leaves a house at a rate of 60 mi/h. At
the same time, another car leaves the same house at a rate of
50 mi/h in the opposite direction. At what time will the cars be
330 miles apart?
Rate Time Distance
First Car 60mph
t
60t
Second Car 50mph
t
50t
60t  50t  330
Motion Problem Set Ups
8. It takes 1 hour longer to fly to St. Paul at 200 mi/h than it
does to return at 250 mi/h. How far away is St. Paul?
Rate
Time Distance
Fly to St. Paul
200mph t  1 200  t  1
Rtn from St. Paul 250mph
t
250t
200 t 1  250t
Tuesday Night’s Homework
Solve Each of today’s worksheet problems
Wednesday Warm Up
The Charlestown Chiefs football team scored
10 times for a total of 46 points. If a
touchdown is worth 7 points and a field goal is
worth 3 points, how many touchdowns and
how many field goals did the Chiefs score?
# of Touchdowns: x
# of Field Goals: 10  x
# of Touchdowns: 4
# of Field Goals: 6
7  x   310  x   46
7 x  30  3x  46
4 x  30  46
4 x  16
x4
Two jets leave Dallas at the same time and fly in opposite
directions. One is flying west 50 mi/h faster than the other.
After 2 hours, they are 2500 miles apart. Find the speed of
each jet.
How are the distances that the two jets fly related
Distance flown by 1st Jet + Distance flown by 2nd jet = 2500
mi
Two jets leave Dallas at the same time and fly in opposite
directions. One is flying west 50 mi/h faster than the other.
After 2 hours, they are 2500 miles apart. Find the speed of
each jet.
How are the times that the two jets fly related
The time of flight of 1st Jet = The time of flight of 2nd Jet
Two jets leave Dallas at the same time and fly in opposite
directions. One is flying west 50 mi/h faster than the other.
After 2 hours, they are 2500 miles apart. Find the speed of
each jet.
How are the speeds of the two jets fly related
Speed of Westbound Jet = Speed of Eastbound Jet + 50 mi/hr
Two jets leave Dallas at the same time and fly in opposite
directions. One is flying west 50 mi/h faster than the other.
After 2 hours, they are 2500 miles apart. Find the speed of
each jet. Define: Let x = the speed of the jet flying east.
Then x + 50 = the speed of the jet flying west.
Relate: eastbound jet’s plus westbound jet’s equals the total
distance
distance
distance
Jet
Eastbound
Rate
x
Westbound x + 50
Write:
2 x
Time Distance Traveled
2
2x
2
2(x + 50)
+
2( x + 50 )
Eastbound
Westbound
=
2500
Two jets leave Dallas at the same time and fly in opposite
directions. One is flying west 50 mi/h faster than the other.
After 2 hours, they are 2500 miles apart. Find the speed of
each jet.
Two jets leave Dallas at the same time and fly in opposite
directions. One is flying west 50 mi/h faster than the other.
After 2 hours, they are 2500 miles apart. Find the speed of
each jet.
Jet
Rate
Eastbound
x
Westbound x + 50
Time Distance Traveled
2
2x
2
2(x + 50)
2 x  2 x  100  2500
4 x  100  2500
Eastbound
Westbound
Westbound Jet: 650mph
4 x  2400
x  600mph
Eastbound Jet: 600mph
1. Two trains left Mooseport at the same time. One
traveled north at 83 mph. The other traveled south at 67
mph. After how many hours were the two trains 600
miles apart?
Rate
Northbound
Southbound
Time
Distance
1. Two trains left Mooseport at the same time. One
traveled north at 83 mph. The other traveled south at 67
mph. After how many hours were the two trains 600
miles apart?
Northbound
Southbound
Rate
83 mph
67 mph
Time
Distance
1. Two trains left Mooseport at the same time. One
traveled north at 83 mph. The other traveled south at 67
mph. After how many hours were the two trains 600
miles apart?
Northbound
Southbound
Rate
83 mph
67 mph
Time
t
t
Distance
1. Two trains left Mooseport at the same time. One
traveled north at 83 mph. The other traveled south at 67
mph. After how many hours were the two trains 600
miles apart?
Northbound
Southbound
Rate
83 mph
67 mph
Time
t
t
Distance
83t
67t
83t  67t  600
2. Romeo first saw Juliet when she was 87 meters away.
He started running toward her at a rate of 5 m/s. Three
seconds later, Juliet saw Romeo and began running
toward him at a rate of 4 m/s. How many seconds after
Romeo first saw Juliet did they meet?
Rate
Time
Distance
Romeo
Juliet
2. Romeo first saw Juliet when she was 87 meters away.
He started running toward her at a rate of 5 m/s. Three
seconds later, Juliet saw Romeo and began running
toward him at a rate of 4 m/s. How many seconds after
Romeo first saw Juliet did they meet?
Rate
Time
Distance
Romeo
5 m/s
Juliet
4 m/s
2. Romeo first saw Juliet when she was 87 meters away.
He started running toward her at a rate of 5 m/s. Three
seconds later, Juliet saw Romeo and began running
toward him at a rate of 4 m/s. How many seconds after
Romeo first saw Juliet did they meet?
Rate
Time
Distance
Romeo
5 m/s
t
Juliet
4 m/s
t –3
2. Romeo first saw Juliet when she was 87 meters away.
He started running toward her at a rate of 5 m/s. Three
seconds later, Juliet saw Romeo and began running
toward him at a rate of 4 m/s. How many seconds after
Romeo first saw Juliet did they meet?
Rate
Time
Distance
Romeo
5 m/s
t
5t
Juliet
4 m/s
t –3
4(t – 3)
5t  4 t  3  87
3. Bad Bart is fleeing the scene of a bank robbery at 70
mph. Thirty minutes after he leaves, a police helicopter
leaves the scene to catch him, traveling 100 mph along
the same route. How many hours will Bart have been
traveling when the police catch up?
Rate
Time
Distance
Bad Bart
Police Helo
3. Bad Bart is fleeing the scene of a bank robbery at 70
mph. Thirty minutes after he leaves, a police helicopter
leaves the scene to catch him, traveling 100 mph along
the same route. How many hours will Bart have been
traveling when the police catch up?
Rate
Time
Distance
Bad Bart
70 mph
Police Helo 100 mph
3. Bad Bart is fleeing the scene of a bank robbery at 70
mph. Thirty minutes after he leaves, a police helicopter
leaves the scene to catch him, traveling 100 mph along
the same route. How many hours will Bart have been
traveling when the police catch up?
Rate
Time
Distance
Bad Bart
70 mph
t
Police Helo 100 mph
t – .5
3. Bad Bart is fleeing the scene of a bank robbery at 70
mph. Thirty minutes after he leaves, a police helicopter
leaves the scene to catch him, traveling 100 mph along
the same route. How many hours will Bart have been
traveling when the police catch up?
Rate
Time
Distance
Bad Bart
70 mph
t
70t
Police Helo 100 mph
t – .5 100(t – .5)
70t  100 t  .5
4. Karma rode her bike up a mountain trail at an average
speed of 4 mph. Then she rode back down the trail at an
average speed of 20 mph. The entire trip took 3 hours.
How far up the mountain did Karma go?
Rate
Uphill
Downhill
Time
Distance
4. Karma rode her bike up a mountain trail at an average
speed of 4 mph. Then she rode back down the trail at an
average speed of 20 mph. The entire trip took 3 hours.
How far up the mountain did Karma go?
Uphill
Downhill
Rate
4 mph
20 mph
Time
Distance
4. Karma rode her bike up a mountain trail at an average
speed of 4 mph. Then she rode back down the trail at an
average speed of 20 mph. The entire trip took 3 hours.
How far up the mountain did Karma go?
Uphill
Downhill
Rate
4 mph
20 mph
Time
t
3-t
Distance
4. Karma rode her bike up a mountain trail at an average
speed of 4 mph. Then she rode back down the trail at an
average speed of 20 mph. The entire trip took 3 hours.
How far up the mountain did Karma go?
Uphill
Downhill
Rate
4 mph
20 mph
Time
t
3-t
Distance
4t
20(3 - t )
4t  20 3  t 
Wednesday Night’s Homework
Complete a D = RT table for each problem.
Then use the table to construct an equation.
Thursday Warm Up
Solve each of the problems from last night’s
homework.
1. Two camels pass each other in the desert, going in
opposite directions. One camel is walking at an average
rate of 9 km/h. The other camel is walking at an average
rate of 7 km/h. After how many hours will the camels be
60 km apart?
Rate
Time
Distance
Camel #1
Camel #2
1. Two camels pass each other in the desert, going in
opposite directions. One camel is walking at an average
rate of 9 km/h. The other camel is walking at an average
rate of 7 km/h. After how many hours will the camels be
60 km apart?
Rate
Time
Distance
Camel #1
9 km/hr
Camel #2
7 km/hr
1. Two camels pass each other in the desert, going in
opposite directions. One camel is walking at an average
rate of 9 km/h. The other camel is walking at an average
rate of 7 km/h. After how many hours will the camels be
60 km apart?
Rate
Time
Distance
Camel #1
9 km/hr
t
Camel #2
7 km/hr
t
1. Two camels pass each other in the desert, going in
opposite directions. One camel is walking at an average
rate of 9 km/h. The other camel is walking at an average
rate of 7 km/h. After how many hours will the camels be
60 km apart?
Rate
Time
Distance
Camel #1
9 km/hr
t
9t
Camel #2
7 km/hr
t
7t
9t  7t  60
2. Two camels pass each other in the desert, going In
opposite directions. The rate of one camel is 3 km/h
faster than the rate of the other. Four hours later, the
camels are 68 km apart. Find the rate of the faster camel.
Rate
Camel #1
Camel #2
Time
Distance
2. Two camels pass each other in the desert, going In
opposite directions. The rate of one camel is 3 km/h
faster than the rate of the other. Four hours later, the
camels are 68 km apart. Find the rate of the faster camel.
Camel #1
Camel #2
Rate
r km/hr
(r + 3)km/hr
Time
Distance
2. Two camels pass each other in the desert, going In
opposite directions. The rate of one camel is 3 km/h
faster than the rate of the other. Four hours later, the
camels are 68 km apart. Find the rate of the faster camel.
Camel #1
Camel #2
Rate
r km/hr
(r + 3)km/hr
Time
4 hr
4 hr
Distance
2. Two camels pass each other in the desert, going In
opposite directions. The rate of one camel is 3 km/h
faster than the rate of the other. Four hours later, the
camels are 68 km apart. Find the rate of the faster camel.
Camel #1
Camel #2
Rate
r km/hr
(r + 3)km/hr
Time
4 hr
4 hr
Distance
4r
4(r +3)
4r  4  r  3  68
3. A plane left Emerald City flying at an average rate of
270 mph. Two hours later, another plane left Emerald
City flying in the same direction at 450 mph. How long
will the second plane be flying until it catches up with
the first?
Rate
Time
Distance
1st Plane
2nd Plane
3. A plane left Emerald City flying at an average rate of
270 mph. Two hours later, another plane left Emerald
City flying in the same direction at 450 mph. How long
will the second plane be flying until it catches up with
the first?
Rate
Time
Distance
1st Plane
270 mph
2nd Plane
450 mph
3. A plane left Emerald City flying at an average rate of
270 mph. Two hours later, another plane left Emerald
City flying in the same direction at 450 mph. How long
will the second plane be flying until it catches up with
the first?
Rate
Time
Distance
1st Plane
270 mph
t
2nd Plane
450 mph
t–2
3. A plane left Emerald City flying at an average rate of
270 mph. Two hours later, another plane left Emerald
City flying in the same direction at 450 mph. How long
will the second plane be flying until it catches up with
the first?
Rate
Time
Distance
1st Plane
270 mph
t
270t
2nd Plane
450 mph
t–2
450(t – 2)
270t  450 t  2
4. At 9:00 AM, a jet leaves St. Louis and travels east at
420 mph. At 11:00 AM, another jet leaves St. Louis and
travels west at 350 mph. What time will the two jets be
3535 miles.
Rate
Eastbound
Westbound
Time
Distance
4. At 9:00 AM, a jet leaves St. Louis and travels east at
420 mph. At 11:00 AM, another jet leaves St. Louis and
travels west at 350 mph. What time will the two jets be
3535 miles.
Eastbound
Westbound
Rate
420 mph
350 mph
Time
Distance
4. At 9:00 AM, a jet leaves St. Louis and travels east at
420 mph. At 11:00 AM, another jet leaves St. Louis and
travels west at 350 mph. What time will the two jets be
3535 miles.
Eastbound
Westbound
Rate
420 mph
350 mph
Time
t
t– 2
Distance
4. At 9:00 AM, a jet leaves St. Louis and travels east at
420 mph. At 11:00 AM, another jet leaves St. Louis and
travels west at 350 mph. What time will the two jets be
3535 miles.
Eastbound
Westbound
Rate
420 mph
350 mph
Time
t
t – 2
Distance
420t
350(t – 2)
420t  450 t  2  3535
Thursday Homework
Study for uniform motion quiz by doing
practice problems