Explainer: Applications on Systems of Linear Equations

In this expaliner, we will learn how to translate a real-life problem into a system of equations and to find its solution(s).

The key skill when answering this type of question is to accurately derive the system of equations and use sensible variables that are not misleading. The best way to learn this skill is by looking at examples. Let us do this now.

Example 1: Forming and Solving Systems of Equations

A sorority held a bake sale to raise money and sold brownies and chocolate chip cookies. They priced the brownies at $1 and the chocolate chip cookies at $0.75. They raised $700 and sold 850 items. How many brownies and how many cookies were sold?

Answer

Firstly, we are trying to find the number of brownies and cookies sold, so let 𝑏 be the number of brownies and 𝑐 be the number of cookies. We are told that a brownie costs $1 and a cookie $0.75 and that the total sales are $700. Using this information, we can form an equation for the total sales: 1𝑏+0.75𝑐=700.

If the coefficient of a variable is 1, we tend not to write it, so we can write this as

𝑏+0.75𝑐=700.(1)

We are also told that a total of 850 items have been sold. This means that we can form a second equation:

𝑏+𝑐=850.(2)

Now that we have a system of two equations, we can solve this using whichever method we prefer. Here, we will use elimination. If we subtract (1) from (2), we get 0.25𝑐=150.

If we then divide through by 0.25, we find that 𝑐=600. We then need to substitute this back into one of our equations to find 𝑏. Here, we will substitute into (2) as it is easier: 𝑏+(600)=850.

If we then subtract 600 from both sides of the equation, we find that 𝑏=250.

We have found that the sorority sold 250 brownies and 600 cookies.

Example 2: Solving Story Problems Using Systems of Equations

Daniel bought 3 muffins and 2 cookies for $3.30, while Jacob bought 2 muffins and 5 cookies for $5.50. Work out the price of a single muffin and a single cookie.

Answer

Here, we are asked to calculate the price of a cookie and a muffin from the information we are given. Let us call the price of a muffin π‘š and the price of a cookie 𝑐. Daniel’s purchase then allows us to form the equation 3π‘š+2𝑐=3.30, and Jacob’s purchase allows us to form the equation 2π‘š+5𝑐=5.50.

We then have a system of linear equations that we can solve using standard methods. Here, we are going to use elimination. If we multiply the first equation through by 2 and the second equation through by 3, we get the following system: 6π‘š+4𝑐=6.60,6π‘š+15𝑐=16.50.

If we then subtract the top equation from the bottom equation, we get 11𝑐=9.90.

Dividing through by 11, we find that 𝑐=0.90.

To find 𝑏, we can then substitute this into either of our equations. Here, we will substitute into the first equation: 3π‘š+2(0.90)=3.303π‘š+1.80=3.30.

If we then subtract 1.80 from each side and divide through by 3, we find that π‘š=0.50.

Now we have that a cookie costs $0.90 and a muffin costs $0.50.

Example 3: Solving Real-World Problems Using Systems of Equations

A store clerk sold 60 pairs of sneakers. The high-tops sold for $98.99 and the low-tops sold for $129.99. If the receipts for the two types of sales totaled $6,404.40, how many of each type of sneaker were sold?

Answer

Here, we are trying to find the number of each type of sneakers sold. That is, high- and low-top sneakers. If we then let the number of high-tops sold be β„Ž and the number of low-tops be 𝑙, given that we know the cost of each pair and the total cost of all sales, we can form the equation 98.99β„Ž+129.99𝑙=6,404.40.

We also know that a total of 60 pairs were sold, which gives us the equation β„Ž+𝑙=60.

We now have a system of equations that can be solved. Given that the coefficients of the first equation contain decimals and the second equation is comparatively simple, let us solve this system using substitution. Rearranging the second equation to make β„Ž the subject by subtracting 𝑙, we have β„Ž=60βˆ’π‘™.

If we then substitute this into our first equation, we get 98.99(60βˆ’π‘™)+129.99𝑙=6,404.40.

Expanding the parentheses, we get 5,939.40βˆ’98.99𝑙+129.99𝑙=6,404.40.

Simplifying, we get 5,939.40+31𝑙=6,404.40, and then subtracting 5,939.40 from each side, we get 31𝑙=465.

Finally, we divide through by 31 to get 𝑙=15.

If we then substitute this into our rearranged formula, we find that β„Ž=60βˆ’(15)=45.

Example 4: Solving Word Problems Using Systems of Equations

In a test with 20 questions, π‘₯ marks are awarded for each correct answer and 𝑦 marks are deducted for each incorrect answer. Noah answered 12 questions correctly and 8 questions incorrectly, and he scored 44 points. Chloe answered 14 questions correctly and 6 questions incorrectly, and she scored 58 points. How many points were deducted for each incorrect answer?

Answer

In this question, we are trying to find the number of marks that were deducted for an incorrect answer. Reading the question carefully, we can establish that this is the value of 𝑦. We can use the information about Noah’s test to form the equation 12π‘₯βˆ’8𝑦=44.

We can use Chloe’s test information to form the equation 14π‘₯βˆ’6𝑦=58.

We now have a system of two equations that we can solve using standard methods. Here, we will use elimination. If we multiply the first equation through by 3 and the second equation through by 4, our system becomes 36π‘₯βˆ’24𝑦=132,56π‘₯βˆ’24𝑦=232.

If we then subtract the first equation from the second equation to eliminate 𝑦, we get 20π‘₯=100, and dividing through by 20, we find that π‘₯=5.

To find 𝑦, we then need to substitute this into one of our equations. We will substitute into the first equation, giving us 12(5)βˆ’8𝑦=44, which simplifies to 60βˆ’8𝑦=44.

We can then add 8𝑦 to each side and subtract 44 to get 8𝑦=16.

Dividing through by 8, we find that 𝑦=2.

Example 5: Solving Story Problems Using Systems of Equations

Michael buys 5 apples and 3 bananas from a grocery store and pays $3.40. Natalie buys 3 apples and 2 bananas from the same store and pays $2.10. Work out the price of a single apple and a single banana.

Answer

Here, let us call the price of a banana 𝑏 and the price of an apple π‘Ž. From Michael’s shopping, we can derive the equation 5π‘Ž+3𝑏=3.40, and from Natalie’s shopping, we can derive the equation 3π‘Ž+2𝑏=2.10.

We then have a system that we can solve. Here, we will solve this using elimination by first multiplying the equations to make the coefficients of 𝑏 equal. If we multiply the first equation through by 2 and the second equation through by 3, we get 10π‘Ž+6𝑏=6.80,9π‘Ž+6𝑏=6.30.

Subtracting the second equation from the first, we find that π‘Ž=0.50.

If we then substitute this into our first equation, we can find 𝑏: 5(0.50)+3𝑏=3.40.

Simplifying, we get 2.50+3𝑏=3.40, and subtracting 2.50 from each side and dividing through by 3, we get 𝑏=0.30.

This means that an apple costs $0.50 and a banana costs $0.30, or, in cents, 50Β’ and 30Β’ respectively.

Key Points

To solve story problems, we can use the following method:

  1. Choose variables to represent the unknowns in the problem.
  2. Using these variable, translate the words into mathematical equations.
  3. Solve the equations.
  4. Express your answer in terms of the unknowns from the problem, not in terms of the variables we introduced to solve the problem.

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