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Solving Word Problems Involving System of Equations

Updated on April 27, 2011

SOLVING WORD PROBLEMS INVOLVING SYSTEM OF EQUATIONS

Many problems that we encounter in applications involve more than one variable. When the relations between the variables involve several equations, then we must solve the equations of the resulting system simultaneously.

In this hub I present three interesting word problems that involve solving system of equations, with their corresponding solution. I hope you will find this hub enjoyable.

Problem No. One:

The sum of the digits of a three-digit number is 13. If the tens and hundreds digits are interchanged, the new number is 90 less than the original, and if the units and hundreds digits are interchanged, the resulting number is 99 less than the original. Find the original number.

Solution to Problem Number One:

Representation:

Let X = the unit digit

Y = the tens digit

Z = the hundreds digit

x + 10y + 100z = the original number

Since the problem involves three variables, we must derive three working equations.

Working equation Number One:

Based from the sum of the digits is 13:

X + y + z = 13

Working equation Number Two:

Based from the tens and hundreds digits interchanged resulting to new number

90 less than the original

X + 10z + 100y = x + 10y + 100z - 90

Simplify the equation we get

-90z + 90y = - 90

-z + y = -1 = equation two

Working equation Number Three:

Based from the units and hundreds digits are interchanged, resulting to new number 99 less than the original.

Z + 10y + 100x = x + 10y + 100z - 99

Simplify this equation we get:

-99x + 99x = - 99

-z + x = - 1 equation 3

Our system of equation now consists of

X + y + z = 13 eq. 1

-z + y = - 1 eq. 2

-z + x = -1 eq. 3


Solving equation 1 and equation 2 using elimination method by addition to eliminate z

X + y + z = 13

-z + y = - 1

X + 2y = 12 equation 4

Solving eqn 1 and eqn 3 using elimination method by addition to eliminate z :

X + y + z = 13

-z + x = -1

Y + 2x = 12 - eqn. 5

We now have two new equations

X + 2y = 12 - eqn 4

2x + y = 12 - eqn. 5

Multiply eqn. 4 by two

(x + 2y = 12) 2 --- 2x + 4y = 24

Solving eqn 4 and 5 using elimination method by subtraction to eliminate x

2x + 4y = 24

2x + y = 12

2x + 4y = 24

- 2x + y = -12

(1/3) 3y = 12 )1/3)

y = 4

Substituting y = 4 to either eqn 4 or 5 to solve for x

2x + 4 = 12

-y = -4

(1/2) 2x = 8 (1/2)

X = 4

Substituting x = 4, y = 4 to corresponding variables in equation 1

4 + 4 + z = 13

8 + z = 13

z = 13 - 8

z = 5

The original number is 544.

Problem Number two:

Twenty five coins, whose value is $2.75 are made up of nickel, dime and quarters. If the nickel were dimes, and dimes were quarters, and the quarters nickels, the total value would be $3.75. How many coins each type are there:

Solution:

Representation

Let n = number of nickels

d = number of dimes

q = number of quarters

Working equation number one:

Since there are 25 coins mentioned, the first working equation will be

n + d + q = 25 - eqn. 1

Working equation number two:

Is based on the value of money of $2.75

(.05) n + (.10) d + (.25) q = 2.75

Multiplying the whole equation by 100

5n + 10 d + 25q = 275 ------- eqn. 2

Working equation number three:

Is based on interchanging nickels and dimes, dimes and quarters, quarters and nickels giving new value of $.375.

(.10) n + (.25) d + (.05) q = 3.75

10n + 25d + 5q = 375 …. Eqn. 3

Our system of equation now consists of

n + d + q = 25 --…eqn. 1

5n + 10d + 25q = 275 … eqn. 2

10n + 25d + 5q = 375 ….. eqn. 3

Solving eqn 1 and eqn 2:

Multiplying eqn 1 by 5

(n + d + q = 25)5

Solving eqn 1 and 2 using elimination method by subtraction

5n + 5d + 5q = 125

-5n - 10d - 25q = -275

- 5d - 20q = 150

Multiply -5d - 20q = -150 by -1

5d + 20q = 150 --- dividing the whole eqn by 5 we get d + 4q = 30

Let this be equation 4

Solving eqn 2 and 3

Multiplying eqn 2 by 2

(5n +10d + 25q = 275) 2

10\n + 20d + 50q = 550

Solving eqn 2 and 3 using elimination method by subtraction

10n + 20d + 50 = 5550

-10n - 25d - 5q = -375

- 5d +45q = 175

Dividing the whole equation

-5d + 45q = 175 x 5

We get -d + 9q = 35 --- let this be equation 5

Adding equation 4 and 5

d + 4q = 30 …..eqn. 4

-d + 9q = 35 … eqn. 5

(1/13) 13q = 65 (`1/13)

q = 5

Substitute q = 5 in either egn 4 0r 5

d + 4(5) = 30

d + 20 = 30

d = 20 - 30 = 10

d = 10

Substituting 1 = 5 , d = 1- in egn. 1

n + 10 + 5 = 25

n + 15 = 25 …. n = 25 - 15 = 10

n = 10

There are 5 quarters, 10 dimes and 10 nickels.

Check: (5) (.25) + (10) (.10) + (10) (.05) = 2.75

1.25 + 1.00 + .50 = 2.75

2.75 = 2.75

Problem Number Three:

If A, B and C work together on a job it will take 1-1/3 hours. If only A and B work, it would take 1-5/7 hours, but B and C work, it would take 2-2/5 hours. How long would it take each man, working done, to complete the job.

Solution:

Representation:

Let x = number of hours it will take workers A to complete a job.

Let y = number of hours it will take worker B to complete a job.

Let z = number of hours it will take worker C to complete a job.

Working equation one:

(1/x + 1/y + 1/z) (4/3) = 1

(4/3x + 4/3y + 4/3z = 1) 3 x yz

4yz + 4 x z + 4 x y = 3 x yx = eqn. 1

Working equation number twoL1/x + 1/y) 12/7 = 1

(12/7x + 12/7y = 1) 7 x y

12y + 12z = 7xy ---- eqn. 2

Working equation number three:

(1/y + 1/z) 12/5 = 1

(12/5y + 12/5z =1) 5 yz

12y + 12z = 5yz ….. eqn. 3

Our system of equation now consist of:

4yz + 4xz + 4xy = 3xyz …..eqn. 1

12y + 12x = 7xy …. eqn. 2

12z + 12y = 5yz …. eqn. 3

Using equation 3 to solve for z

12z - 5zy = -12y

z (12-5y)/12-5y = -12y/12-5y

z = -12y/ 12-5y

Substitute this value of z in eqn 1.

4y (-12y/12-5y) + 4x (-12y/12-5y) + 4 x y = 3 xy (-12y/12-5y)

(-48y2 /12-5y - 48x y /12-5y + 4x y = -36y2x/ 12-5y) (12-5y)

-48y2 - 48 x y + 4 x y (12-5y) = 36y2 x

-48y2 - 48 x y + 48 x y - 20 xy2 = - 36 y2 x

-48 y2 = -36y2x + 20 y2 x

(-48y2 = -16 y2 x ) -1/16 y2 )

3 = x

x = 3

Substituting x = 3 to eqn. 2 to solve for y

12 y + 12(3) = (7)(3)y

12y + 36 = 21y

36 = 21y - 12y

36/9 = 9y/9 y = 4

Substituting y = 4, in eqn. 3 to solve for z

12z + 12(y ) = 5(4)z

12z + 48 = 20z

48 = 20z - 12 z

1/8 (48) = 8z (1/8)

Z = 6

It will take 3 hours for worker A, 4 hours for worker B and 6 hours for worker C to complete the job.

Check: (1/3 + ¼ + 1/6) 4/3 = 1

4/9 + 4/12 + 4/18 = 1

16/36 + 12/36 + 8/ 36 = 1

36 / 36 = 1

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    • profile image

      kimberly 5 years ago

      nice, it help me to understand, how to solve coin problem easier

    • profile image

      Teisha Marie Sunga 5 years ago

      some person find math difficult but when you study very well in math you can easily identify it so work hard to make it perfect.I am only 12 years old and I am a a grade 6 student of the Philippines and here in the Philippines education is hard and it is fun and also educatoin is high here in the Philippines.

    • profile image

      Noel Jaramillo 6 years ago

      nice to have you to review my kids. They have exams in relations, functions, variations and sequences, series, and you're here to help, thanks so much

    • profile image

      Dexter Simons 6 years ago

      ty for that article..

      you help me in my assignment.

    • profile image

      M Shahid Habib 7 years ago

      Its a nice technique but you can solve these problems with the help of matrices

    • cristina327 profile image
      Author

      cristina327 7 years ago from Manila

      Hi Dave it is nice to hear from you again. Thank you for visiting this hub. Regards.

    • Dave Mathews profile image

      Dave Mathews 7 years ago from NORTH YORK,ONTARIO,CANADA

      W O W ! Never knew that Ate' Cristina was so smart in Math. W O W !

      KUYA Dave.

    • cristina327 profile image
      Author

      cristina327 7 years ago from Manila

      Hi Pastor_Walt it is nice to hear from you again. I am happy that you find the problems presented here enjoyable. Your visit is greatly appreciated.Remain blessed always. Best regards.

    • Pastor_Walt profile image

      Pastor_Walt 7 years ago from Jefferson City, Tennessee

      I've always been a math "nut." Enjoyable probs. High School algebra was always my favorite class.

      Maranatha