Oct 31, 2009

Linear Equations Word Problems


Word problems for systems of linear equations are troublesome for most of the students in understanding the situations and bringing the word problem into equations. We tried to explain the trick of solving word problems for equations with two variables with an example.
Example:
2000 tickets were sold in an exhibition on Saturday. The cost of a ticket for an adult is $4 and for a child is $2. The total amount collected on Saturday was $6400. Find the number of adult tickets and child tickets sold on Saturday.
Solution:
Step 1:
Total number of tickets sold = 2000
Cost of ticket for an adult = $4
Cost of ticket for a child = $2
Total amount collected = $6400
We are asked to find the number of adult tickets and children tickets sold.
Let x be the number of adult tickets sold and y the number of child tickets sold on Saturday.
Step 2:
Total number of tickets sold = 2000
Number of adults tickets sold + Number of child tickets sold = 2000
x + y = 2000 --------------- (1)
Total amount collected = Amount collected from adult tickets + Amount collected from child tickets
Since there are x adults and the adult ticket cost is $4 per, the amount collected from adult tickets is 4x and since there are y children and the child ticket costs $2 per child, the amount collected from child tickets is 2y.
6400 = 4x + 2y
 It can be written as 4x + 2y = 6400 ---------------- (2)
Now, we got a system of two linear equations in two variables.
x + y = 2000 --------------- (1)
4x + 2y = 6400 ---------------- (2)
Step 3:
Solve the above two linear equations to find the value of x and y.
Multiply equation (1) with -2 and add the resulting equation and equation (2) to eliminate the variable y.
     -2x - 2y = -4000
(+) 4x + 2y = 6400
--------------------------------
       2x = 2400                                    
Divide by 2 on both the sides and simplify.
x = 1200
Step 4:
Now, substitute the value of x = 1200 in either of the two equations.
Let us substitute in equation (1). So, it becomes
1200 + y = 2000
Subtract 1200 from both sides of the equation and simplify.
1200 + y - 1200 = 2000 - 1200
y = 800
Step 5:
So, the solution for the given system of equations is (1200, 800), which means 1200 adult tickets and 800 child tickets were sold on Saturday. 

Note: The above problem can be also solved using substitution method since the coefficients of x and y in the first equation is 1.

 Practice Problems:
1. James bought 5 apples and 10 oranges for $4. Donald bought 3 apples and 9 oranges for $3. The shop keeper strictly told that there will not be any discounts. What is the cost of an apple and an orange?
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2. Ana writes test to upgrade her level. The test has 25 questions for a total score of 150 points. Among the 25 questions, each multiple choice questions carries 3 points and the descriptive type questions carries 8 points. How many multiple choice questions and descriptive type questions are there in the test?
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3. The sum of the digits of a two digit number is 7. When the digits are reversed, then number is decreased by 9. Find the number.
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4. The perimeter of a rectangle is 12 meters. The length is 3 more than twice its width. Find the dimensions of the rectangle.(This can also be solved just with one variable)
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5. The sum of two numbers is 12. When three times the first number is added to 5 times the second number, the resultant number is 44. Find the two numbers.
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Having problem in solving any of these questions? Post your doubt in comment. We will help you how to solve this.
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27 comments:

  1. Hi , I need help with Practice Problem #2 (Ana and her test questions...).
    Really look forward to some help!
    What method are you using to solve this?
    thanks!
    ReplyDelete
  2. Number of MCQ = x
    Total points for MCQ = 3x
    Number of descriptive type = y
    Total points for descriptive = 8y
    Total questions = Number of MCQ + Number of descriptive
    25 = x + y ………(This is equation 1)
    Total points = Total points for MCQ + Total points for descriptive
    150 = 3x + 8y ……..(This is equation 2)
    Solve the equations to find x and y.
    You can use either substitution or elimination method.
    ReplyDelete
  3. i just don't get how to do it
    ReplyDelete
  4. hello :)
    what are the two equations for #4, please? thanks.
    ReplyDelete
  5. Question 4:
    Length = 5m; Width = 1m
    Length = x
    Width = y
    Perimeter = 2x + 2y
    2x + 2y = 12
    => x + y = 6 …(1)
    Also
    x = 2 + 2y
    x – 2y = 2 …(2)

    I hope this can help you.
    ReplyDelete
  6. #5 what is the solutiion
    ReplyDelete
  7. Click on the "show answer" to find the solution at the bottom of each question.
    ReplyDelete
  8. do you have more harder questions that the ones above
    ReplyDelete
  9. Hi , I need help with Practice Problem #1 Really look forward to some help!
    What method are you using to solve this?
    thanks!
    ReplyDelete
  10. cost of an apple = x
    cost of 5 apples = 5x
    cost of 3 apples = 3x

    cost of an orange = y
    cost of 10 oranges = 10y
    cost of 9 oranges = 9y

    James bought 5 apples and 10 oranges for $4.
    Translation into algebraic expression,
    5x + 10y = 4 ....(1)

    Donald bought 3 apples and 9 oranges for $3.
    Translation into algebraic expression,
    3x + 9y = 3 ....(2)

    Multiply eqn(1) by 3 and eqn(2) by 5
    15x + 30y = 12
    15x + 45y = 15

    - - -
    ________________
    -15y = -3
    y = -3/-15 = 1/5 = 0.20
    Cost of an orange = $0.20 or 20 cents
    Now, plugin the value of y in (1) or (2) to solve for x.

    Note: Elimination method is used here.
    ReplyDelete
  11. solution of question#1. plz :)
    ReplyDelete
  12. Cost of an apple = $0.40 or 40 cents
    Cost of an orange = $0.20 or 20 cents
    ReplyDelete
  13. Hello,I need help with Practice Problem #3.Can you plz help me??
    ReplyDelete
  14. Before solving question no 3, let us practically analyze this problem. What is the sum of digits of a number 23? 2+3 = 5. When it gets reversed, the number becomes 32. Difference between 23 and 32 is 9. 23 can be written as 20+3 and 32 can be written as 30+2. Suppose x and y are the digits. Sum is supposed to be x + y = 7 .....(1). When gets reversed, yx (or 10y+x) is 9 more than xy(or 10x+y).
    10x + y +9 = 10y+x
    9x-9y = -9
    dividing by 9 on both sides
    x-y = -1 .....(2)
    Solve (1) and (2) to find the value of x and y.
    ReplyDelete
  15. HI I am having a hard time trying to make an equation out of this word problem. a man goes into a grocery store and buys a grapefruit and a toy ball. A 7% tax is applied to the price of non-food items. If the bill is $5.15, including $0.18 in tax on the toy ball, what are the prices of the grapefruit and the ball? I just need to figure out how to set this up as an equation. I am stumped
    ReplyDelete
  16. 7% tax is applied to the price of non-food items. Here, toy ball is non-food item. So, 7% tax is applied to the cost of toy ball.

    Let the cost of toy ball is x.
    Tax on toy ball = $0.18
    tax = 7% of cost = 0.18
    0.07 of cost = 0.18
    0.07x = 0.18
    Here I am confused. I dont get terminating decimals. Please check out the numbers. If my guess is correct, it would be 6% tax.
    In that case, 0.06x = 0.18, solving for x, we have x = $3.
    Cost of toy = $3 + $0.18= $3.18
    Now, you know how to find the cost of grape fruit. Good luck.
    ReplyDelete
  17. Please help me with question 5.
    ReplyDelete
  18. x = first number
    y = second number
    sum = x + y = 12 ...(1)

    3 times first number = 3x
    5 times second number = 5y
    sum = 3x + 5y = 44 ...(2)

    Solve the equations and find the solution.
    ReplyDelete
  19. Number 4 please? Help me how to solve for it. Thanks!
    ReplyDelete
  20. A LENDING library has a fixed charge for the first three days and an additional charge for each day thereafter.Saritha paid Rs 27 for a book kept for seven days,while Susy paid Rs 21 for the book she kept for five days. find the fixed charge and the charge for each extra day.
    ReplyDelete
  21. Help me solve linear equation please. this is for our math project.
    ReplyDelete
  22. how do i solve

    x+5y=1
    2x+3y=9

    i tried using addition, subtraction, multiplication and division but i couldn't seem to get rid of a variable
    ReplyDelete
  23. x + 5y = 1 ....(1) x 2
    2x + 3y = 9 ....(2) x 1

    After multiplication, we have

    2x + 10y = 2
    2x + 3y = 9 (subtract)
    ____________
    7y = -7
    Dividing by 7 on both sides, we have y = -7. Substitute the value of y in any of the equations and find the value of x.
    ReplyDelete
  24. The website is superb!
    ReplyDelete
  25. hello.......Could u pls help me with the 1st practice question?
    ReplyDelete
  26. Hello Shravya,

    It is already explained in one of my reply.
    ReplyDelete

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