3 5linear word problems ii

Linear Word-Problems II
Frank Ma © 2011

Following are key-words translated into mathematic operations.

+: add, sum, plus, total, combine, increased by, # more than ..

–: subtract, difference, minus, decreased by, # less than ..

*: multiply, product, times, “fractions or %” of the amount ..

/: divide, quotient, shared equally, ratio ..

Twice = Double = 2* (amount)

Square = (amount)2

Square = (amount)2
In chapter 2, we solved problems with a single variable.

Square = (amount)2
In chapter 2, we solved problems with a single variable. Many of those problems may be solved using two variables and a system of equations.

Square = (amount)2
In chapter 2, we solved problems with a single variable. Many of those problems may be solved using two variables and a system of equations. The advantage of using two variables instead of one is that it's easier to construct the equations to form the system.

Square = (amount)2
In chapter 2, we solved problems with a single variable. Many of those problems may be solved using two variables and a system of equations. The advantage of using two variables instead of one is that it's easier to construct the equations to form the system. To do this:
I. identify the unknown quantities and named them as x and y

Square = (amount)2
I. identify the unknown quantities and named them as x and y II. usually there are two numerical relations given, translate
each numerical relation into an equation to make a system

Square = (amount)2
I. identify the unknown quantities and named them as x and y II. usually there are two numerical relations given, translate
each numerical relation into an equation to make a system
III. solve the system.

Example A. A 30-foot rope is cut into two pieces. The longer one is 6 feet less than twice of the short piece. How long is each piece?

Let x = length of the long piece.
y = length of the short piece
30
x
y

Then x + y = 30 --- E1
30
x
y

Then x + y = 30 --- E1
x = 2y – 6 --- E2
30
x
y
{

Then x + y = 30 --- E1
x = 2y – 6 --- E2
30
x
y
{
Use the substitution method.

Then x + y = 30 --- E1
x = 2y – 6 --- E2
30
x
y
{
Substitute x = (2y – 6) into E1,

Then x + y = 30 --- E1
x = 2y – 6 --- E2
30
x
y
{
Substitute x = (2y – 6) into E1, we get
(2y – 6) + y = 30

Then x + y = 30 --- E1
x = 2y – 6 --- E2
30
x
y
{
(2y – 6) + y = 30 solve for y
2y – 6 + y = 30

Then x + y = 30 --- E1
x = 2y – 6 --- E2
30
x
y
{
(2y – 6) + y = 30 solve for y
2y – 6 + y = 30
3y – 6 = 30

Then x + y = 30 --- E1
x = 2y – 6 --- E2
30
x
y
{
(2y – 6) + y = 30 solve for y
2y – 6 + y = 30
3y – 6 = 30
3y = 36

Then x + y = 30 --- E1
x = 2y – 6 --- E2
30
x
y
{
(2y – 6) + y = 30 solve for y
2y – 6 + y = 30
3y – 6 = 30
3y = 36
y = 36/3 = 12

Then x + y = 30 --- E1
x = 2y – 6 --- E2
30
x
y
{
(2y – 6) + y = 30 solve for y
2y – 6 + y = 30
3y – 6 = 30
3y = 36
y = 36/3 = 12
Put y = 12 into E2, we get x = 2(12) – 6

Then x + y = 30 --- E1
x = 2y – 6 --- E2
30
x
y
{
(2y – 6) + y = 30 solve for y
2y – 6 + y = 30
3y – 6 = 30
3y = 36
y = 36/3 = 12
Put y = 12 into E2, we get x = 2(12) – 6 =18.

Then x + y = 30 --- E1
x = 2y – 6 --- E2
30
x
y
{
(2y – 6) + y = 30 solve for y
2y – 6 + y = 30
3y – 6 = 30
3y = 36
y = 36/3 = 12
Put y = 12 into E2, we get x = 2(12) – 6 =18.
Hence the two pieces are 18ft and 12ft.

Example B. Suppose four hamburgers and three fries costs $18, and three hamburgers and two fries costs $13. Find the price of each.

Let x = price of a hamburger, y = price of fries

4x + 3y = 18
The system is

E1
4x + 3y = 18
3x + 2y = 13
{
The system is
E2

E1
4x + 3y = 18
3x + 2y = 13
{
The system is
E2
Use the elimination method.

E1
4x + 3y = 18
3x + 2y = 13
{
The system is
E2
Use the elimination method. LCM of 3y and 2y is 6y.

E1
4x + 3y = 18
3x + 2y = 13
{
The system is
E2
Multiply E1 by 2
8x + 6y = 36
2*E1:

E1
4x + 3y = 18
3x + 2y = 13
{
The system is
E2
Multiply E1 by 2 and E2 by 3,
8x + 6y = 36
9x + 6y = 39
2*E1:
3*E2:

E1
4x + 3y = 18
3x + 2y = 13
{
The system is
E2
Multiply E1 by 2 and E2 by 3, subtract
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:

E1
4x + 3y = 18
3x + 2y = 13
{
The system is
E2
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
–x = –3  x = 3

E1
4x + 3y = 18
3x + 2y = 13
{
The system is
E2
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
–x = –3

E1
4x + 3y = 18
3x + 2y = 13
{
The system is
E2
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
–x = –3  x = 3
To find y, put x = 3 into E2,

E1
4x + 3y = 18
3x + 2y = 13
{
The system is
E2
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
–x = –3  x = 3
To find y, put x = 3 into E2, 3(3) + 2y = 13

E1
4x + 3y = 18
3x + 2y = 13
{
The system is
E2
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
–x = –3  x = 3
9 + 2y = 13

E1
4x + 3y = 18
3x + 2y = 13
{
The system is
E2
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
–x = –3  x = 3
9 + 2y = 13
2y = 13 – 9

E1
4x + 3y = 18
3x + 2y = 13
{
The system is
E2
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
–x = –3  x = 3
9 + 2y = 13
2y = 13 – 9
2y = 4 y = 4/2 = 2.

E1
4x + 3y = 18
3x + 2y = 13
{
The system is
E2
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
–x = –3  x = 3
9 + 2y = 13
2y = 13 – 9
2y = 4 y = 4/2 = 2.
So a hamburger is $3, and fries are $2.

Following problems use formulas in setting up the equations.

The RTD Formula

The RTD Formula
Let R = Rate or speed (usually mph)
T = Time (in hours)
D = Distance (in miles)
then RT = D.

The RTD Formula
T = Time (in hours)
then RT = D.
For example, if we drove from LA to SF at a rate of 75 mph and it took 6 hours,

The RTD Formula
T = Time (in hours)
then RT = D.
For example, if we drove from LA to SF at a rate of 75 mph and it took 6 hours, it means R = 75, T = 6.

The RTD Formula
T = Time (in hours)
then RT = D.
For example, if we drove from LA to SF at a rate of 75 mph and it took 6 hours, it means R = 75, T = 6. Therefore
75(6) = 450 miles is the distance between the two cities.

The RTD Formula
T = Time (in hours)
then RT = D.
Up-and-down-stream-problems

The RTD Formula
T = Time (in hours)
then RT = D.
When a boat travels up and down a stream, the speed of the boat depends on the boat's speed R in still water (cruising speed)

The RTD Formula
T = Time (in hours)
then RT = D.
When a boat travels up and down a stream, the speed of the boat depends on the boat's speed R in still water (cruising speed) and the current's speed C.

The RTD Formula
T = Time (in hours)
then RT = D.
When a boat travels up and down a stream, the speed of the boat depends on the boat's speed R in still water (cruising speed) and the current's speed C. Going upstream, the current slows the boat down

The RTD Formula
T = Time (in hours)
then RT = D.
When a boat travels up and down a stream, the speed of the boat depends on the boat's speed R in still water (cruising speed) and the current's speed C. Going upstream, the current slows the boat down and going downstream it speeds up the boat.

The RTD Formula
T = Time (in hours)
then RT = D.
When a boat travels up and down a stream, the speed of the boat depends on the boat's speed R in still water (cruising speed) and the current's speed C. Going upstream, the current slows the boat down and going downstream it speeds up the boat. In fact, the upstream speed is (R – C),

The RTD Formula
T = Time (in hours)
then RT = D.
When a boat travels up and down a stream, the speed of the boat depends on the boat's speed R in still water (cruising speed) and the current's speed C. Going upstream, the current slows the boat down and going downstream it speeds up the boat. In fact, the upstream speed is (R – C), and the downstream speed is (R + C).

For example, a boat that travels 3 mph in still water goes upstream against a 1 mph current,

For example, a boat that travels 3 mph in still water goes upstream against a 1 mph current, it's upstream rate is 2mph.

It would take 6 hours to go 12 miles.

It would take 6 hours to go 12 miles. If it travels downstream,
it can travel at 3 mph

it can travel at 4 mph and it would take 3 hours to go 12 miles.

Example C. A boat can travel upstream for 24 miles in 6 hours and the same distance downstream in 3 hours.
What is the cruising speed of the boat and the current speed?

Let R = cruising speed, C = current speed,

Make a table.

Make a table.
By the formula R*T = D we get two equations.

{
6(R – C) = 24
3(R + C) = 24
--- E1
--- E2

{
6(R – C) = 24
3(R + C) = 24
--- E1
--- E2
Simplify each equation, divide E1 by 6 and divide E2 by 3.

{
6(R – C) = 24
3(R + C) = 24
--- E1
--- E2
{
R – C = 4 --- E3
R + C = 8 --- E4

{
6(R – C) = 24
3(R + C) = 24
--- E1
--- E2
{
R – C = 4 --- E3
R + C = 8 --- E4
Add the equations to eliminate C
+)

{
6(R – C) = 24
3(R + C) = 24
--- E1
--- E2
{
R – C = 4 --- E3
R + C = 8 --- E4
+)
2R = 12

{
6(R – C) = 24
3(R + C) = 24
--- E1
--- E2
{
R – C = 4 --- E3
R + C = 8 --- E4
+)
2R = 12
R = 6

{
6(R – C) = 24
3(R + C) = 24
--- E1
--- E2
{
R – C = 4 --- E3
R + C = 8 --- E4
+)
2R = 12
R = 6
Substitute R = 6 into E2,

{
6(R – C) = 24
3(R + C) = 24
--- E1
--- E2
{
R – C = 4 --- E3
R + C = 8 --- E4
+)
2R = 12
R = 6
6 + C = 8
C = 2

{
6(R – C) = 24
3(R + C) = 24
--- E1
--- E2
{
R – C = 4 --- E3
R + C = 8 --- E4
+)
2R = 12
R = 6
6 + C = 8
C = 2
R = cruising speed = 6 mph, C = current speed = 2 mph
So,

{
6(R – C) = 24
3(R + C) = 24
--- E1
--- E2
{
R – C = 4 --- E3
R + C = 8 --- E4
+)
2R = 12
R = 6
6 + C = 8
C = 2
So,
Simple Interest Formula

{
6(R – C) = 24
3(R + C) = 24
--- E1
--- E2
{
R – C = 4 --- E3
R + C = 8 --- E4
+)
2R = 12
R = 6
6 + C = 8
C = 2
So,
Recall the simple interest of an investment is I = R*P

{
6(R – C) = 24
3(R + C) = 24
--- E1
--- E2
{
R – C = 4 --- E3
R + C = 8 --- E4
+)
2R = 12
R = 6
6 + C = 8
C = 2
So,
Recall the simple interest of an investment is I = R*P where
I = amount of interest for one year
R = interest (in %)
P = principal

Example D. (Mixed investments)
We have $20,000 saved in two accounts which give 5% and
6% interest respectively. In one year, their combined
interest is $1150. How much is in each account?

Make a table using the interest formula R*P = I

6
*x
100
5
*y
100

6
*x
100
5
*y
100
{
x + y = 20,000 --- E1
6
5
y = 1150 --- E2
x +
100
100

6
*x
100
5
*y
100
{
x + y = 20,000 --- E1
6
5
y = 1150 --- E2
x +
100
100
Multiply E2 by 100 to clear the denominator.

6
*x
100
5
*y
100
{
x + y = 20,000 --- E1
6
5
y = 1150 --- E2
x +
100
100
6
5
(
y = 1150)100
x +
100
100

6
*x
100
5
*y
100
{
x + y = 20,000 --- E1
6
5
y = 1150 --- E2
x +
100
100
100
6
5
(
y = 1150)100
x +
100
100

6
*x
100
5
*y
100
{
x + y = 20,000 --- E1
6
5
y = 1150 --- E2
x +
100
100
100
6
5
(
y = 1150)100
x +
100
100
6x + 5y = 115000 ---E3

{
x + y = 20,000 ---E1
6x + 5y = 115000 ---E3

{
x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
To eliminate y, multiply E1 by -5
-5x – 5y = -100,000
E1(-5):

{
x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
To eliminate y, multiply E1 by -5 and add to E3.
-5x – 5y = -100,000
E1(-5):
6x + 5y = 115000
+)

{
x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
-5x – 5y = -100,000
E1(-5):
6x + 5y = 115000
+)
x = 15000

{
x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
-5x – 5y = -100,000
E1(-5):
6x + 5y = 115000
+)
x = 15000
Sub x=15000 into E1,
15000 + y = 20000

{
x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
-5x – 5y = -100,000
E1(-5):
6x + 5y = 115000
+)
x = 15000
15000 + y = 20000
y = 5000

{
x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
-5x – 5y = -100,000
E1(-5):
6x + 5y = 115000
+)
x = 15000
15000 + y = 20000
y = 5000
So there are $15,000 in the 6% account,
and $5,000 in the 5% account.

3 5linear word problems ii

by math123a on Jun 22, 2011

Accessibility

Categories

Tags

Upload Details

Usage Rights

Statistics

3 5linear word problems ii — Presentation Transcript