Lesson HOW TO algebraize and solve these problems on 2 equations in 2 unknowns

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HOW TO algebraize and solve these problems on 2 equations in 2 unknowns

     In standard/typical word problems solved using systems of two equations, creating a system of equations is not, as a rule, a difficult task.

     But in some cases you need to stretch your mind to do it correctly . . . 

     This is why I placed this lesson and these problems here.


     Although the problems talk about THREE unknown quantities, they admit reducing to 2 equations in 2 unknowns.

     I wanted, most of all, to avoid having 3 equations, and I succeeded in it.

     Now I want to familiarize you with this approach/technique.

Problem 1

A stadium has 50,000 seats. Seats sell for $25 in section A, $20 in section B, $15 in section C.
The number of seats in section A equals the total number of seats in sections B and C.
Suppose the stadium takes in $1,074,500 from each sold-out event. How many seats does each section holds?

Solution

Let x be the number of seats in section B, and y be the number of seats in section C.


Then the number of seats in section A is x+y, the sum of seats in sections B and C, according to the condition.


The condition says

seatsA + seatsB + seatsC = 50000,   or

(x+y)  + x      + y      = 50000,   which is the same as

2(x+y) = 50000,                     which implies

x + y =  = 25000.


Thus we found that the sum of seats in sections B and C is 25000,  

and, therefore, the number of seats in section A is 25000.


Thus we have now the problem  FOR TWO UNKNOWNS only, since we just excluded A.

Now our system of two equations is

  x +   y = 25000,                  (1)
20x + 15y = 1074500 - 25*25000.     (2)


Equation (2) express the money payed for seats in sections B and C together.


Simplify (1),(2) step by step:

  x +   y = 25000,                  (3)
20x + 15y = 449500.                 (4)


Multiply eq(1) by 20 (both sides). Then subtract eq(2) from it:

20x + 20y = 500000,                 (3)
20x + 15y = 449500.                 (4)
-------------------------

====> 5y = 500000-449500 = 50500  ====>  y =  = 10100.


Then  x = 25000 - y = 25000 - 10100 = 14900.


Answer.  There are 25000 sears in section A,  14900 in section B  and  10100 in section C.

Problem 2

Bruce collects stamps. He has six times as many 10-cent stamps as 5-cent stamps, and he has some 20 cent stamps as well.
Bruce has 72 stamps with a total value of $8.40. How many of each stamp does he have?

Solution

Bruce has

(x of 5-cent stumps) + (6x 10-cent stamps) + (y 20-cent stumps)


Hence, you have these two equations:

(1)  x  + 6x      +    y =  72           (counting stamps)
(2)  5x + 10*(6x) + 20*y = 840  cents    (counting stamps' values) 


Simplify the equations

(1)   7x +   y =  72,
(2)  65x + 20y = 840.


From (1), express  y = 72-7x  and substitute it into eq(2). You will get

65x + 20*(72-7x) = 840  ====>

65x + 1440 - 140x = 840  ====>

-75x = 840 - 1440 = -600  ====>  x =  = 8.


So, Bruce has  8  5-cent stamps;  8*6 = 48  10-cent stamps  and  72 - (8+48) = 16    20-cent stamps.


Answer.  Bruce has  8  5-cent stamps;  48  10-cent stamps  and  16  20-cent stamps.


Check.  5*8 + 10*48 + 20*16 = 840.  ! Correct !

Problem 3

Sally is going to buy a total of 11 new items at Target. She is going to buy jeans, dresses, and shoes.
She is going to spend exactly $460 and has discovered that jeans are $25, dresses are $50, and shoes are $40.
She is also going to buy twice as many shoes as jeans. Find out how many jeans, how many shoes, and how many dresses she will buy?

Solution

Let x = the unknown number of jeans, and

let y = the unknown number of dresses.

Then the unknown number of shoes is 2x, according to the condition.


Now we can write our equations, TWO equations in TWO unknowns:


  x +     2x +    y =  11,     (1)    (counting for items)

25x + 40*(2x) + 50y = 460.     (2)    (counting dollars of spending)



Simplify the system:

  3x +   y =  11,              (3)
105x + 50y = 460.              (4)


To solve the system, I will apply the substitution method. 

For it, express y = 11-3x from (3) and substitute it into eq(4). You will get


105x + 50*(11-3x) = 460  ====>

105x + 550 - 150x = 460  ====>  -45x = 460 - 550 = -90  ====>  x =  = 2.

Thus we found the unknown x.  It is x= 2.


Then from eq(3)  y = 11 - 3x = 11- 3*2 = 11-6 = 5.


Answer.  2 jeans, 2*2 = 4 shoes  and  5 dresses.


Check.   2*25 + 4*40 + 5*50 = 460.    ! Correct !

Problem 4

Jack and Jill spent two weeks touring Boston, New York City, Philadelphia, and Washington D.C.
They paid $120, $200, $80, and $100 per night respectively. Their total bill was $2020.
The number of days spent in NYC was the same as the sum of the days spent in Boston and D.C.
They spent three times as many days in NYC as they did in Philly. How many days did they stay in each city?

Solution

Usual practice in hotels is to pay for night.

But, specially for the purpose of this problem, I will assume that they pay for each DAY and that the number of days was 14 ("two weeks" literally).

Solution

Let P be the number of days in Philly.

Then the number of days in NYC is 3P, from the condition.

We also are given that the number of days in NYC, 3P, is equal to B + W.


So we have

P + 3P + (B + W) = 14,   or,  replacing  B + W  by "3P",  

P + 3P + 3P = 14,

7P = 14  ====>  P =  = 2.


Answer.  2 days in P;  6 days in NYC;  and remaining  14 - (2+6) = 6 days in  B + W.

It is how my logic works.
Still, it is not the final solution yet, but I just reduced the problem to the only 2 (two, TWO) unknowns B and W,

and at this point, I am sure, now you are able to complete the solution WITHOUT MY HELP,
using REMAINING CONDITION on the total cost.

Problem 5

Two cars, A and B, 148 kilometers apart, traveled toward each other. A started at 6 a.m. while B started an hour later.
They met on the road at 9 a.m. Had both of them started at 6 a.m., they would have met at 8:28. Find the speeds.

Solution

1.  In the first scenario, 1-st car was on the way 3 hours (from 6 am till 9 am) before meeting.

                           2-nd car was on the way 2 hours before meeting.


    Let "a" and "b" be their rates in miles per MINUTE, respectively. 

    Then from the first scenario, the total distance between the cities was

        180a + 120b  = 148  miles.               (1)    (3 hours = 180 minutes;  2 hours = 120 minutes !)



2.  In the second scenario, each car was on the way 2 hours and 28 minutes = 148 minutes.  


    Therefore, the second equation for the entire distance is

        148a + 148b = 148  miles,   or

           a + b = 1       (miles per minute)     (2) 



3.  Thus you have this system of two equations in two unknowns

       a +    b =   1,    and

    180a + 120b = 148.


    Solve it by ANY method you want.  Both methods - Substitution and Elimination - are good.  Choose any and complete the solution on your own.

Problem 6

John and Nat were given some money. If John spends $50 and Nat spends $100 each day, John would still have $2500 left
while Nat would have spent all her money. If John spends $100 and Nat spends $50 each day, John would still have $1000 left
while Nat would have spent all her money. How much money John and Nat were given each?

Solution

Let J be the amount John had initially and let x be the number of days in the 1-st scenario.

Let N be the amount Nat  had initially and let y be the number of days in the 2-nd scenario.


For the first scenario, we have these two equations

    J = 50x + 2500,     (1)

    N = 100x.           (2)


For the second scenario, we have these two equations

    J = 100y + 1000,     (3)

    N = 50y.             (4)


In all, we have 4 equations in 4 unknown  J, N, x and y, so we have a chance to solve it.



From eq(2), express x =   and substitute it into eq(1).  You will have

    J = 50*(N/100) + 2500,   or  J = (1/2)*N + 2500,   or   2J - N = 5000  (5)


From eq(4), express y =   and substitute it into eq(3).  You will have

    J = 100*(N/50) + 1000,    or  J = 2N + 1000,       or    J -2N = 1000  (6)



Thus, we reduced the problem to two equations in 2 unknowns

    2J -  N = 5000    (5)

     J - 2N = 1000    (6)


From eq(6), express  J = 2N + 1000  and substitute it into eq(5).  You will get

    2*(2N + 1000) - N = 5000,


which implies

    4N + 2000 - N = 5000

    3N            = 5000 - 2000 = 3000

     N            = 3000/3 = 1000.


Then from eq(5)

    2J = 5000 + 1000 = 6000

     J = 6000/2 = 3000.


ANSWER.  John had 3000;  Nat had 1000.

Problem 7

At first, the ratio of Bob's and Ted's money was 5 to 2. After each of them spent an equal amount,
the ratio of bob to teds became 6 to 1. At the end, they had a total 126 dollars. How much did they spend together?

Solution

From the condition, Bob and Ted had initially  5x and 2x dollars, respectively.


Then each of them spent equal amount of "y" dollars.


After that, they  possessed  5x-y  and  2x-y dollars respectively, and  


     =  = 6,

so

    5x-y = 6*(2x-y),

    5x - y = 12x - 6y

     5y = 7x.            (1)


Also, we are given that

    (5x-y) + (2x-y) = 126  dollars,   or


     7x - 2y = 126.      (2)


In (2), replace  7x  by  5y, based on (1).  You will get


    5y - 2y = 126,

    3y      = 126

     y      = 126/3 = 42.


Then from (1),  x =  =  = 5*6 = 30.


So, initially they had  5x + 2x = 7x = 7*30 = 210 dollars.   


Hence, they spent  210 - 126 = 84 dollars.    ANSWER

Problem 8

May spent 1/6 of her money on a dress and 2 blouses. The dress costs as much as 3 blouses.
She spends 3/4 of the remaining money on a watch. The watch costs $220.50 more than the dress.
How much did she have at first.

Solution

Let x = How much she had at first (dollars).


She first spent 1/6 of her money - so,    of her amount remained.


Taking into account that the dress costs as much as 3 blouses, we can write the system of equations in this form

     = 3B + 2B                (1)

     = 3B + 220.50     (2)


Simplify

     = 5B                     (3)

     = 3B + 220.50         (4)


Multiply equation (3) by 6 (both sides).  Multiply equation (4) by 24 (both sides).  You will get

    x = 30*B                         (5)

    15x = 72B + 5292                 (6)


From (5), substitute the expression for x into (6). You will get

    15*30*B = 72*B + 5292

    450*B - 72*B = 5292

    378*B = 5292

        B = 5292/378 = 14.


Then from (5) we get the FINAL ANSWER:  

    x = 30*14 = 420 dollars.


ANSWER.  She had originally  420 dollars.

Problem 9

Manuel bought 6 candy bars and 3 sodas at the gas station for $8.40. Gary bought
3 candy bars and 4 sodas at the same gas station for $7.45. What is the price of one soda?

Solution

Write equations as you read the problem


    6x + 3y = 840   cents

    3x + 4y = 745   cents


I will apply the determinant method (= the Cramer's rule) to find y, the soda's price.


    y =  = 130.


ANSWER.  The soda's price is $1.30.

Problem 10

Mrs. Lee bought 20 ducks and chickens altogether for 126 dollars. Each chicken costs 2 dollars less than each duck.
If she bought 6 more chickens than ducks, how much did she pay for each chicken ?

Solution

I will solve the problem in two steps.


In  STEP 1,  I will determine the number of chickens (x) and ducks (y).

For it, I write two equations from the condition


    x + y = 20      (1)    (total)

    x - y =  6      (2)    (the difference).


To find x, add two equations


    2x = 26  ---->  x = 26/2 = 13.


Then from equation (2),  y = 20-x = 20-13 = 7.


So, 13 chicken and 7 ducs were bought.




Now in  STEP 2,  I will solve for the price of each chicken (p).


For it, I write the total money equation


    13p + 7*(p+2) = 126.


From this equation


    13p + 7p + 14 = 126

    20p = 126 - 14 = 112.


From this equation,   p = 112/20 = 5  dollars = 5   dollars = 5.60 dollars = 5 dollars and 60 cents.    ANSWER

Problem 11

Daniel has one-peso coins in the left and right pockets of his pants.
If he transfers one coin from his left pocket to his right pocket,
the number of coins in his right pocket would be twice of what he has on the left.
However, if he transfers one coin from his right to his left, the number of coins
in his pockets will be equal. How many coins does he have in his right pocket?

Solution

Let R be the number of one-peso coins in his right pocket; 

let L be the number of one-peso coins in his left pocket.


Write equations as you read the problem


    2*(L - 1) = R + 1    (1)

    L + 1 = R - 1        (2)


From the second equation, L = R-2.   Substitute it into the first equation and find R


    2*((R-2) - 1) = R + 1,

    2R - 4 - 2 = R + 1

    2R - R  = 1 + 4 + 2

       R    =   7


ANSWER.  There are 7 coins in the right pocket.

Problem 12

A tortoise makes a journey in two parts; it can either walk at 4 cm/s or crawl at 3 cm/s.
If the tortoise walks the first part and crawls the second, it takes 110 seconds.
If it crawls the first part and walks the second, it takes 100 seconds.
Find the lengths of the two parts of the journey.

Solution

Let x be the length of the 1st part of the journey, and

let y be the length of the 2nd part of the journey.


In the first  scenario, the travel time is   + .

In the second scenario, the travel time is   + .


Therefore, from the problem's description we have this system of two equations in two unknowns

     +  = 110  seconds    (1)

     +  = 100  seconds    (2)


To simplify it, multiply equations (1) and (2) by 12 (both sides).  You will get then

    4x  +  3y = 1320    (3)

    3x  +  4y = 1200    (4)


Solve it using the Elimination method. 
Multiply equation (3) by 3; multiply equation (4) by 4.  You will get

    12x +  9y = 3960    (5)

    12x + 16y = 4800    (6)


Subtract eq(5) from eq(6).   You will get

          16y - 9y = 4800 - 3960

              7y   =  840

               y   =  840/7 = 120.


Next find x, substituting y= 120 to equation (3).

    4x + 3*120 = 1320  -->  4x = 1320 - 3*120  -->  4x = 960  -->  x = 960/4 = 240.


ANSWER.  First part of the journey is 240 cm.  Second part of the journey is 120 cm.


CHECK.  Make checking on your own by substituting the found values of x an d to equations (1) and (2).

Problem 13

The monthly salaries of two men are in the ratio 3:2
and their expenditures are in the ratio 8:5.
Each man saves 500 dollars every month. Find their monthly salaries.

Solution

From the problem, their salaries     are 3x and 2x, where x is some common measure (now unknown).

                  Their expenditures are 8y and 5y, where y is some common measure (now unknown).


Write equations as you read the problem

    3x - 8y = 500     (1)

    2x - 5y = 500     (2)


Solve equations using the Elimination method. 
For it, multiply equation (1) by 2; multiply equation (2) by 3.  You will get

    6x - 16y = 1000    (1)

    6x - 15y = 1500    (2)


Now subtract equation (1) from equation (2).  You will get

           y = 500.


Next, from equation ((1) you get

    3x - 8*500 = 500,  3x = 500 + 4000 = 4500,  x = 4500/3 = 1500.


Thus the salary of the first man is 3x = 3*1500 = 4500;  the salary of the second man is 2x = 2*1500 = 3000.

Problem 14

A farmer has chickens and ducks in his farm.
If he sells 2 chickens and 3 ducks a day, there would be 100 chickens left when all the ducks are sold.
If he sells 3 chickens and 2 ducks a day, there would be 25 chickens left after all the ducks are sold.
Find the number of ducks and chickens on the farm.

Solution

Let C be the number of chickens and D be the number of ducks on the farm.


In the first scenario, the number of days is ; so, according to the 
problems description, in the first scenario

    C -  = 100.    (1)


In the second scenario, the number of days is ; so, according to the 
problems description, in the second scenario

    C -  =  25.    (2)


Now subtract equation (2) from equation (1)  (both sides).  You will get

     -  = 75.


Multiply both sides by 6

      9D  -  4D = 450

         5D     = 450

          D     = 450/5 = 90.


Then from equation (1)

    C =  + 100 = 60 + 100 = 160.


ANSWER.  160 chickens and 90 ducks.

Problem 15

A factory makes use of two basic machines, A and B, which turn out two different products, yarn and thread.
Each unit of yarn requires 1 hour on machine A and 2 hours on machine B, while each unit of thread requires 1 hour on A and 1 hour on B.
Machine A runs 8 hours per day, while machine B runs 14 hours per day.
How many units each of yarn and thread should the factory make to keep its machines running at capacity?

Solution

Let Y be the number of units of yarn, and

let T be the number of units of thread.


Write equations as you read the problem

    1*Y + 1*T =  8   hours   (machine A)     (1)

    2*Y + 1*T = 14   hours   (machine B)     (2)


     +--------------------------------+
     |   Thus the setup is complete.  |
     +--------------------------------+


To find Y, subtract eq(1) from eq(2).   You will get

    2Y - Y    = 14 - 8 = 6,  Y = 6.


Now find T from equation (1)

    T = 8 - Y = 8 = 6 = 2.


ANSWER.  6 units of yarn and 2 units of thread.

My other lessons in this site on solving systems of two linear equations in two unknowns (Algebra-I curriculum) are
    - Solution of the linear system of two equations in two unknowns by the Substitution method
    - Solution of the linear system of two equations in two unknowns by the Elimination method
    - Solution of the linear system of two equations in two unknowns using determinant
    - Geometric interpretation of the linear system of two equations in two unknowns
    - Useful tricks when solving systems of 2 equations in 2 unknowns by the Substitution method
    - Solving word problems using linear systems of two equations in two unknowns

    - Word problems that lead to a simple system of two equations in two unknowns
    - Oranges and grapefruits
    - Using systems of equations to solve problems on tickets
    - Three methods for solving standard (typical) problems on tickets
    - Using systems of equations to solve problems on shares
    - Using systems of equations to solve problems on investment
    - Two mechanics work on a car
    - The Robinson family and the Sanders family each used their sprinklers last summer
    - Roses and vilolets
    - Counting calories and grams of fat in combined food
    - A theater group made appearances in two cities
    - Exchange problems solved using systems of linear equations
    - Typical word problems on systems of 2 equations in 2 unknowns
    - One unusual problem to solve using system of two equations
    - Non-standard problem with a tricky setup
    - Sometimes one equation is enough to find two unknowns in a unique way
    - Solving mentally word problems on two equations in two unknowns
    - Solving systems of non-linear equations by reducing to linear ones
    - Solving word problems for 3 unknowns by reducing to equations in 2 unknowns
    - System of equations helps to solve a problem for the Thanksgiving day
    - Using system of two equations to solve the problem for the day of April, 1

    - OVERVIEW of lessons on solving systems of two linear equations in two unknowns

Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I.

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