SUBSTITUTION METHOD WORD PROBLEMS AND ANSWERS
Problem 1 :
The coach of a cricket team buys 7 bats and 6 balls for $3800. Later, she buys 3 bats and 5 balls for $1750. Find the cost if each bat and each ball.
Let "x" be the cost of each bat.
Let "y" be the cost of each ball.
7x + 6y = 3800 -----(1)
3x + 5y = 1750 -----(2)
Solve (1) for y.
6y = 3800 - 7x
y = (3800 - 7x)/6 -----(3)
Substitute y = (3800 - 7 x)/6 in (2)
(2)-----> 3x + 5(3800 - 7x)/6 = 1750
[18x + 5(3800 - 7x)]/6 = 1750
(18x + 19000 - 35x)/6 = 1750
-17x + 19000 = 1750(6)
-17x + 19000 = 10500
-17x = 10500 - 19000
-17x = -8500
x = 8500/17
x = 500
Substitute x = 500 in (3)
(3)-----> y = [3800 - 7(500)] / 6
y = (3800 - 3500) / 6
y = 300/6
y = 50
So, the cost of each bat is $500 and each ball is $50.
Problem 2 :
The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is $105 and for a journey of 15 km, the charge paid is $155. What are the fixed charge and charge per km ? How much does a person have to pay for traveling a distance of 25 km ?
Let "x" be the fixed charge
Let "y" be the charge per km for the distance covered
x + 10y = 105 ------(1)
x + 15y = 155 ------(2)
Solving (1) for x.
x = 105 - 10y -----(3)
Substitute x = 105 - 10y in (2).
(2)-----> 105 - 10y + 15y = 155
105 + 5y = 155
5y = 50
y = 10
Substitute y = 10 (3).
(30-----> x = 105 -10(10)
x = 105 - 100
x = 5
Therefore, the fixed charge is $5 and charge per km for the distance covered is $10.
Amount has to be paid for a travel of 25 km is
= 5 + 25(10)
= 5 + 250
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Worksheet on word problems on rational numbers.
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Solving Systems of Equations Real World Problems
Wow! You have learned many different strategies for solving systems of equations! First we started with Graphing Systems of Equations . Then we moved onto solving systems using the Substitution Method . In our last lesson we used the Linear Combinations or Addition Method to solve systems of equations.
Now we are ready to apply these strategies to solve real world problems! Are you ready? First let's look at some guidelines for solving real world problems and then we'll look at a few examples.
Steps For Solving Real World Problems
- Highlight the important information in the problem that will help write two equations.
- Define your variables
- Write two equations
- Use one of the methods for solving systems of equations to solve.
- Check your answers by substituting your ordered pair into the original equations.
- Answer the questions in the real world problems. Always write your answer in complete sentences!
Ok... let's look at a few examples. Follow along with me. (Having a calculator will make it easier for you to follow along.)
Example 1: Systems Word Problems
You are running a concession stand at a basketball game. You are selling hot dogs and sodas. Each hot dog costs $1.50 and each soda costs $0.50. At the end of the night you made a total of $78.50. You sold a total of 87 hot dogs and sodas combined. You must report the number of hot dogs sold and the number of sodas sold. How many hot dogs were sold and how many sodas were sold?
1. Let's start by identifying the important information:
- hot dogs cost $1.50
- Sodas cost $0.50
- Made a total of $78.50
- Sold 87 hot dogs and sodas combined
2. Define your variables.
- Ask yourself, "What am I trying to solve for? What don't I know?
In this problem, I don't know how many hot dogs or sodas were sold. So this is what each variable will stand for. (Usually the question at the end will give you this information).
Let x = the number of hot dogs sold
Let y = the number of sodas sold
3. Write two equations.
One equation will be related to the price and one equation will be related to the quantity (or number) of hot dogs and sodas sold.
1.50x + 0.50y = 78.50 (Equation related to cost)
x + y = 87 (Equation related to the number sold)
We can choose any method that we like to solve the system of equations. I am going to choose the substitution method since I can easily solve the 2nd equation for y.
5. Think about what this solution means.
x is the number of hot dogs and x = 35. That means that 35 hot dogs were sold.
y is the number of sodas and y = 52. That means that 52 sodas were sold.
6. Write your answer in a complete sentence.
35 hot dogs were sold and 52 sodas were sold.
7. Check your work by substituting.
1.50x + 0.50y = 78.50
1.50(35) + 0.50(52) = 78.50
52.50 + 26 = 78.50
35 + 52 = 87
Since both equations check properly, we know that our answers are correct!
That wasn't too bad, was it? The hardest part is writing the equations. From there you already know the strategies for solving. Think carefully about what's happening in the problem when trying to write the two equations.
Example 2: Another Word Problem
You and a friend go to Tacos Galore for lunch. You order three soft tacos and three burritos and your total bill is $11.25. Your friend's bill is $10.00 for four soft tacos and two burritos. How much do soft tacos cost? How much do burritos cost?
- 3 soft tacos + 3 burritos cost $11.25
- 4 soft tacos + 2 burritos cost $10.00
In this problem, I don't know the price of the soft tacos or the price of the burritos.
Let x = the price of 1 soft taco
Let y = the price of 1 burrito
One equation will be related your lunch and one equation will be related to your friend's lunch.
3x + 3y = 11.25 (Equation representing your lunch)
4x + 2y = 10 (Equation representing your friend's lunch)
We can choose any method that we like to solve the system of equations. I am going to choose the combinations method.
5. Think about what the solution means in context of the problem.
x = the price of 1 soft taco and x = 1.25.
That means that 1 soft tacos costs $1.25.
y = the price of 1 burrito and y = 2.5.
That means that 1 burrito costs $2.50.
Yes, I know that word problems can be intimidating, but this is the whole reason why we are learning these skills. You must be able to apply your knowledge!
If you have difficulty with real world problems, you can find more examples and practice problems in the Algebra Class E-course.
Take a look at the questions that other students have submitted:
Problem about the WNBA
Systems problem about ages
Problem about milk consumption in the U.S.
Vans and Buses? How many rode in each?
Telephone Plans problem
Systems problem about hats and scarves
Apples and guavas please!
How much did Alice spend on shoes?
All about stamps
Going to the movies
Small pitchers and large pitchers - how much will they hold?
Chickens and dogs in the farm yard
- System of Equations
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Writing Systems of Linear Equations from Word Problems
Some word problems require the use of systems of linear equations . Here are clues to know when a word problem requires you to write a system of linear equations:
(i) There are two different quantities involved: for instance, the number of adults and the number of children, the number of large boxes and the number of small boxes, etc. (ii) There is a value associated with each quantity: for instance, the price of an adult ticket or a children's ticket, or the number of items in a large box as opposed to a small box.
Such problems often require you to write two different linear equations in two variables. Typically, one equation will relate the number of quantities (people or boxes) and the other equation will relate the values (price of tickets or number of items in the boxes).
Here are some steps to follow:
1. Understand the problem.
Understand all the words used in stating the problem. Understand what you are asked to find. Familiarize the problem situation.
2. Translate the problem to an equation.
Assign a variable (or variables) to represent the unknown. Clearly state what the variable represents.
3. Carry out the plan and solve the problem.
Use substitution , elimination or graphing method to solve the problem.
The cost of admission to a popular music concert was $ 162 for 12 children and 3 adults. The admission was $ 122 for 8 children and 3 adults in another music concert. How much was the admission for each child and adult?
1 . Understand the problem:
The admission cost for 12 children and 3 adults was $ 162 . The admission cost for 8 children and 3 adults was $ 122 .
2 . Translate the problem to an equation.
Let x represent the admission cost for each child. Let y represent the admission cost for each adult. The admission cost for 12 children plus 3 adults is equal to $ 162 . That is, 12 x + 3 y = 162 . The admission cost for 8 children plus 3 adults is equal to $122. That is, 8 x + 3 y = 122 .
3 . Carry out the plan and solve the problem.
Subtract the second equation from the first. 12 x + 3 y = 162 8 x + 3 y = 122 _ 4 x = 40 x = 10 Substitute 10 for x in 8 x + 3 y = 122 . 8 ( 10 ) + 3 y = 122 80 + 3 y = 122 3 y = 42 y = 14 Therefore, the cost of admission for each child is $ 10 and each adult is $ 14 .
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