8 linear word problems in x&y x

Following are key-words translated into mathematic operations.
Linear Word-Problems II

+: 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
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
form the system. To do this:
I. identify the unknown quantities and named them as x and y

Square = (amount)2
II. usually there are two numerical relations given, translate
each numerical relation into an equation to make a system

Square = (amount)2
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?

30 ft

Let L = length of the long piece
S = length of the short piece
30 ft
SL

Then L + S = 30 --- E1 30 ft
SL

Then L + S = 30 --- E1
L = 2S – 6 --- E2{ 30 ft
SL

Then L + S = 30 --- E1
L = 2S – 6 --- E2{
Use the substitution method.
30 ft
SL

Then L + S = 30 --- E1
L = 2S – 6 --- E2{
Substitute L = (2S – 6) into E1, we get
(2S – 6) + S = 30
30 ft
SL

Then L + S = 30 --- E1
L = 2S – 6 --- E2{
(2S – 6) + S = 30
2S – 6 + S = 30
3S – 6 = 30
30 ft
SL

Then L + S = 30 --- E1
L = 2S – 6 --- E2{
(2S – 6) + S = 30
2S – 6 + S = 30
3S – 6 = 30
3S = 36
S = 36/3 = 12
30 ft
SL

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

Then L + S = 30 --- E1
L = 2S – 6 --- E2{
(2S – 6) + S = 30
2S – 6 + S = 30
3S – 6 = 30
3S = 36
S = 36/3 = 12
Put y = 12 into E2, we get x = 2(12) – 6 = 18.
Hence the rope is cut into 18 and 12 feet.
30 ft
SL

Making Tables
Linear Word-Problems
If the given information are the same types of
for multiple entities, organize the information into a table.

Making Tables
Example B. Put the following information into a table.
a. Maria’s grocery list:
6 apples, 4 bananas, 3 cakes.
Don's grocery list:
10 cakes, 2 apples, 6 bananas.

Making Tables
Don's grocery list:
Apple Banana Cake
Maria
Don

Making Tables
Don's grocery list:
Apple Banana Cake
Maria 6 4 3
Don

Making Tables
Don's grocery list:
Apple Banana Cake
Maria 6 4 3
Don 2 6 10

Making Tables
Don's grocery list:
Apple Banana Cake
Maria 6 4 3
Don 2 6 10
b. Beets cost $5/lb and carrots is $3/lb,
Mary bought 8lb of beets and 15Ib of carrots.
Don bought 12Ib of carrots and 9lb of beets .
Beet-cost ($5/lb) Carrot-cost ($3/lb) Total
Maria
Don
Total:

Making Tables
Don's grocery list:
Apple Banana Cake
Maria 6 4 3
Don 2 6 10
Maria (8×5) 40 (15×3) 45 85
Don
Total:

Making Tables
Don's grocery list:
Apple Banana Cake
Maria 6 4 3
Don 2 6 10
Maria (8×5) 40 (15×3) 45 85
Don (9×5) 45 (12×3) 36 81
Total: 166$

Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.

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

Berger
Cost $
Fries
Cost $
Total
Cost $
Joe
Mary

Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary

Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
{
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Use the elimination method.
{
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Use the elimination method. LCM of 3y and 2y is 6y.
{
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Multiply E1 by 2
8x + 6y = 362*E1:
{
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

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

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

4x + 3y = 18
3x + 2y = 13
The system is
E1
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
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
{
–x = –3  x = 3
9 + 2y = 13
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
{
–x = –3  x = 3
9 + 2y = 13
2y = 13 – 9
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

4x + 3y = 18
3x + 2y = 13
The system is
E1
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.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

4x + 3y = 18
3x + 2y = 13
The system is
E1
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 an order of fries is $2.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

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.
and it took 6 hours, it means R = 75, T = 6.

The RTD Formula
T = Time (in hours)
then RT = D.
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.
speed) and the current's speed C.

The RTD Formula
T = Time (in hours)
then RT = D.
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.
current slows the boat down and going downstream it speeds
up the boat.

The RTD Formula
T = Time (in hours)
then RT = D.
up the boat. In fact, the upstream speed is (R – C),

The RTD Formula
T = Time (in hours)
then RT = D.
Following formulas in setting up the equations.
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,

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 4 mph

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

Example D. 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.
Rate Time (T) Distance (D)
Upstream
Downstream

Make a table.
Upstream R – C
Downstream R + C

Make a table.
Upstream R – C 6
Downstream R + C 3

Make a table.
Upstream R – C 6 24
Downstream R + C 3 24

Make a table.
Upstream R – C 6 24
Downstream R + C 3 24
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{
R – C = 4 --- E3
R + C = 8 --- E4
--- E1
--- E2
{

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

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

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

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

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

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

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

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

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

rate principal interest
6%
5%
total
Make a table using the interest formula R*P = I

6% x
5% y
total

6% x
5% y
total
6
100
5
100 * y
* x

6% x
5% y
total 20,000
6
100
5
100 * y
* x

6% x
5% y
total 20,000 1150
6
100
5
100 * y
* x

6% x
5% y
total 20,000 1150
6
100
5
100 * y
* x
x + y = 20,000 --- E1
6
100
5
100
y = 1150 --- E2x +{

6% x
5% y
total 20,000 1150
6
100
5
100 * y
* x
x + y = 20,000 --- E1
6
100
5
100
y = 1150 --- E2x +{
Multiply E2 by 100 to clear the denominator.

6% x
5% y
total 20,000 1150
6
100
5
100 * y
* x
x + y = 20,000 --- E1
6
100
5
100
y = 1150 --- E2x +{
6
100
5
100
y = 1150)100x +(

6% x
5% y
total 20,000 1150
6
100
5
100 * y
* x
x + y = 20,000 --- E1
6
100
5
100
y = 1150 --- E2x +{
6
100
5
100
y = 1150)100x +(
100

6% x
5% y
total 20,000 1150
6
100
5
100 * y
* x
x + y = 20,000 --- E1
6
100
5
100
y = 1150 --- E2x +{
6
100
5
100
y = 1150)100x +(
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.
6x + 5y = 115000
-5x – 5y = -100,000
+)
{
E1(-5):

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

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

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

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

Exercise. A. The problems below are from the word–problems for linear
equations with one variable.
1. A and B are to share $120. A gets $30 more than B, how much does each get?
(Just let A = $ that Mr. A has, and B = $ that Mr. B has.)
2. A and B are to share $120. A gets $30 more than twice of what B gets, how
much does each get?
3. A and B are to share $120. A gets $16 less than three times of what B gets,
how much does each get?
Let x = number of lb of peanuts y = number of lb of cashews (Make tables)
4. Peanuts costs $2/lb, cashews costs $8/lb. How many lbs of each are needed
to get 50 lbs of peanut–cashews–mixture that’s $3/lb?
5. Peanuts costs $3/lb, cashews costs $6/lb. How many lbs of each are needed
to get 12 lbs of peanut–cashews–mixture that‘s $4/lb?
6. We have $3000 more saved at an account that gives 5% interest than at an
account that gives 4% interest. In one year, their combined interest is $600. How
much is in each account?
7. We have saved at a 6% account $2000 less than twice at a 4% account. In
one year, their combined interest is $1680. How much is in each account?
8. The combined saving in two accounts is $12,000. One account gives 5%
interest, the other gives 4% interest. The combined interest is %570 in one year.
How much is in each account?

B. Find x = $ of cashews and y = $ of peanuts per lb, solve for x and y with the
following information.
9. Order 1. (in lb) Order 2 (in lb)
Cashews 2 3
Peanuts 3 1
Cost $16 $17
10.
Order 1. (in lb) Order 2 (in lb)
Cashews 2 3
Peanuts 3 1
Cost $17 $15
11. Order 1 consists of 5 lb of cashews and 3 lb of peanuts and it costs $16,
Order 2 consists of 7 lb of cashews and 2 lb of peanuts and it costs $18.
Organize this information into a table and solve foe x and y.
12. Order 1 consists of 3 lb of cashews and 4 lb of peanuts and it costs $38,
Order 2 consists of 2 lb of cashews and 5 lb of peanuts and it costs $30.
Organize this information into a table and solve for x and y.
Note that 9–12 are the same as the hamburger–fries problem.

13. We have $3000 more saved in an account that gives 5% interest than in an
account that gives 4% interest. In one year, their combined interest is $600. How
much is in each account?
14. We have saved in a 6% account $2000 less than twice than in a 4% account. In
one year, their combined interest is $1680. How much is in each account?
15. The combined saving in two accounts is $12,000. One account has 5% interest,
the other has 4% interest. The combined interest is $570 in one year. How much is
in each account?
16. The combined saving in two accounts is $15,000. One account has 6% interest,
the other has 3% interest. The combined interest is $750 in one year. How much is
in each account?
C. Select the x and y, make a table, then solve.

8 linear word problems in x&y x

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