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Word Problems of Linear Equations By L.D.

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Word Problems of Linear EquationsBy L.D.

Problem 1A wilderness group sells boxes of nuts and boxes of popcorn for a trip to Disneyland. A box of nuts sells for \$4.50 and a box of popcorn sells for \$3. The group makes \$252 and sells twice as many boxes of nuts as popcorn. How many boxes of nuts were sold, how many boxes of popcorn were sold?

Problem 1A wilderness group sells cans of nuts and boxes of popcorn for a trip to Disneyland. A can of nuts sells for \$4.50 and a box of popcorn sells for \$3. The group makes \$252 and sells twice as many boxes of nuts as popcorn as nuts. How many boxes of nuts were sold, how many boxes of popcorn were sold?

Well, to figure that out we need to make an equation set, so highlight all the important information.

Problem 1The first equation is to find out the amount of nuts and popcorn sold due to the quantity that affects the money, so we take the first half of information that was in red and make an equation out of it. A box of nuts sells for \$4.50 and a box of popcorn sells for \$3. The group makes \$252

Problem 1A can of nuts sells for \$4.50 and a box of popcorn sells for \$3. The group makes \$252 (Note: the final amount, what the group makes will always be on the end) So first we take the variable n for the amount of nuts and park that next to the price to get 4.5n Next, we do the same thing with the popcorn and take p for the amount of popcorn and park it next to the price as 3p. Keeping in mind what I said in my note before do you think you can do the equation?

Problem 1The answer is 4.5n + 3p = 252 and that is one of the equations we will use. We will move to the making of the next equation now.

Problem 1For the next equation we need to take the second half of what I originally highlighted in red and turn that into a equation, before I go ahead, try to do it yourself. [the group] sells twice as many boxes of nuts [then they do popcorn]

Problem 1Now using [the group] sells twice as many boxes of nuts [then they do popcorn] to do a equation we take the variables from before p (for the amount of popcorn) and n (for the amount of nuts) and put them into a equation. Since we know the group sells twice as many boxes of nuts as popcorn and twice can be also said as 2, it is safe to say that 2p= n.

Problem 1Now putting both those things together we get the equations 2p = n 4.5n + 3p = 252 Now try to solve these using any method you want, for this I personally prefer substitution.

Problem 1:4.5(p2) + 3p = 252 9p + 3p = 252 12p = 252 12 12 p = 21

2p = n 4.5n + 3p = 252

Problem 1:4.5(p2) + 3p = 252 9p + 3p = 252 12p = 252 12 12 p = 21

2p = n 4.5n + 3p = 252

2 (21) = n 42 = n

Problem 1:4.5(p2) + 3p = 252 9p + 3p = 252 12p = 252 12 12 p = 21

2p = n 4.5n + 3p = 252

The final answer is that they sold 21 boxes of popcorn and 42 boxes of nuts.

2 (21) = n 42 = n

Problem 2You are selling tickets to a high school play. Student tickets cost \$5 and regular tickets cost \$8. You sell 556 tickets and collect \$3797. How many of each type of ticket did you sell?

Problem 2You are selling tickets to a high school play. Student tickets cost \$5 and regular tickets cost \$8. You sell 556 tickets and collect \$3797. How many of each type of ticket did you sell?

Problem 2In this problem we have to figure out the amount of tickets sold and we need to find how that affected the price. We dont know the amount so for the amount of student tickets sold we will use variable s and for general admission we will use the variable g.

Problem 2First we will make an equation for the amount of tickets sold using the variables s and g from before. We know that they got 556 tickets from the equation so to get that you have to add student and general submission together to get. g + s = 556 or to make it easier I will reorder it to be 556 s = g .

Problem 2We still havent found the amount of tickets meaning to the money earned which is \$3797. To do this we need to multiply the amounts or in our case the stand ins (s and g) to the costs and add those to get \$3797. The equation we will get is

Problem 2We still havent found the amount of tickets meaning to the money earned which is \$3797. To do this we need to multiply the amounts or in our case the stand ins (s and g) to the costs and add those to get \$3797. The equation we will get is 5s + 8g = 3797

Problem 2Now its time to find what s and g are we will use our two problem to find out (with substitution). 556 s = g 5s + 8g = 3797 5s + 8(556 s) = 3797 5s + 4448 8s = 3797 -4448 -4448 5s +0 - 8s = -651 5s 8s = -651 3s = -651 3 3

Problem 2Now its time to find what s and g are we will use our two problem to find out (with substitution). 556 - s = g 5s + 8g = 3797 5s + 8(556 s) = 3797 5s + 4448 8s = 3797 -4448 -4448 5s +0 - 8s = -651 5s 8s = -651 -3s = -651 -3 -3 s = 217

Problem 2Now its time to find what s 556 (217) = g 556 217 = g and g are we will use our 339 = g two problem to find out (with substitution). 556 s = g 5s + 8g = 3797

Problem 2So our answer is that there we 217 students that went and 339 general admissions.

Problem 3Two families go to a hockey game. One family purchases two (2) adult tickets and four (4) child tickets for \$28. The other family purchases four (4) adult tickets and five (5) child tickets for \$45.50 dollars. How much does each type of ticket cost?

Problem 3Two families go to a hockey game. One family purchases two (2) adult tickets and four (4) child tickets for \$28. The other family purchases four (4) adult tickets and five (5) child tickets for \$45.50 dollars. How much does each type of ticket cost?

Problem 3The first equation we will make is the one that is the first I highlighted (two (2) adult tickets and four (4) child tickets for \$28). We need to find out the cost of the adult and child tickets so for our problem we will use a and c as variables respectively. We know that it all cost \$28 and that we have 2 adult tickets that cost a and 4 child tickets that cost c. Both those added together equal \$28 so our equation must be 2a + 4c = 28

Problem 3The second equation we will make is the one that is the second highlighted (four (4) adult tickets and five (5) child tickets for \$45.50 dollars). We need to find out the cost of the adult and child tickets so for our problem we will use a and c as variables respectively. We know that it all cost \$45.50 and that we have 4 adult tickets that cost a and 5 child tickets that cost c. Both those added together equal \$45.50 so our equation must be 4a + 5c = 45.5

Problem 3Now to make it so we can substitute we have to take one of the problems and make it equal just c or a. I will do this to the first one. (2a + 4c = 28) 2 a + 2c = 14 -2c -2c

Problem 3Now to make it so we can substitute we have to take one of the problems and make it equal just c or a. I will do this to the first one. (2a + 4c = 28) 2 a + 2c = 14 -2c -2c a = 14 2c

Problem 3Now we take both the problems and substitute to find the cost of c and a. a = 14 2c 4a + 5c = 45.5

4(14 2c) + 5c = 45.5 56 8c + 5c = 45.5 56 - 3c = 45.5 -56 -56 -3c = -10.5 -3 -3

Problem 3Now we take both the problems and substitute to find the cost of c and a. a = 14 2c 4a + 5c = 45.5

4(14 2c) + 5c = 45.5 56 8c + 5c = 45.5 56 - 3c = 45.5 -56 -56 -3c = -10.5 -3 -3 c = 3.5

Problem 3Now we enter the answer we got into a problem. a = 14 2c 4a + 5c = 45.5

a = 14 2(3.5) a = 14 7

Problem 3Now we enter the answer we got into a problem. a = 14 2c 4a + 5c = 45.5

a = 14 2(3.5) a = 14 7 a=7

Problem 3Now we know that adult tickets cost \$7 and childrens tickets cost \$3.50.

AcknowledgementsThank you to Ms. Becker for letting me (L.D.) use her word problems for this. ##### Solving Word Problems - Pearson School · PDF fileSolving Word Problems ... direction for teaching mathematics word problems that will produce success. ... Here is a word problem whose
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