Section 9.2 Models with Systems of Equations
You might have an instict for word problems. If that is the case, great, feel free to jump to the checkpoints below. However, many students are very uncomfortable with word problems. The hard part is translating the sentences into equations. Here is a step-by-step procedure that you might find helpful:
Fact 9.4. Solving Linear Word Problems with Systems.
To solve a word problem about linear functions involving a system of equations, follow the following steps:
Write the information as a list. Our brains have trouble seeing the important information in a paragraph, so writing it as a short list makes it easier to see what you're working with.
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Organize the information into a table like the one below. You can tell what the rows are based on what kinds of totals you are given.
Table 9.5. Sample Table for Solving Systems Models First input units Second input units Total amounts First total units Second total units Units are your friend here, and the variables (the middle two columns) are typically the two quantities you are looking for. The table can be hard to understand in abstract, so I recommend looking carefully at the example below.
Using your table in step 2, write a system of equations. Each row should be an equation, and unless the problem specifies, you can use whatever letters you want for the variables. We recommend choosing letters that have some meaning to what they represent.
Solve the system of equations using either substitution or elimination.
Look back at the question to see what it is asking for. Use your solution to answer the problem's question.
Example 9.6.
Suppose you are selling tickets to an event. There are two kinds of tickets: adult tickets cost $13.50, and child tickets cost $6.75. Suppose on one night you sold 41 tickets and made a total of $398.25. How many of each kind of ticket did you sell?
Step 1:
Adult tickets cost $13.50
Child tickets cost $6.75
Total tickets sold: 41
Total money made: $398.25
Step 2: The two totals we have are money made and tickets sold, so those will be our rows. The two things we don't know are the number of adult tickets sold and number of child tickets sold, so those will be our variables. So, our table looks like this:
| Adult Tickets | Child Tickets | Total amounts | |
| Tickets Sold | Number of Adult Tickets | Number of Child Tickets | Total tickets sold |
| Money Made | Money from Adult Tickets | Money from Child Tickets | Total money made |
Now, we want to fill in numbers to the inside of the table. Since the things we are looking for are the number of adult tickets sold and number of child tickets sold, we will make those our variables. We can pick any letters we want, but it's helpful to pick letters that are connected to what they represent. Let's say \(A\) is the number of adult tickets and \(C\) is the number of child tickets. Therefore, our table looks like this:
| Adult Tickets | Child Tickets | Total amounts | |
| Tickets Sold | \({\color{red}{A}}\) | \({\color{red}{C}}\) | \({\color{red}41}\) |
| Money Made | Money from Adult Tickets | Money from Child Tickets | Total money made |
Now we just have to fill in the last row, and this is where it will be helpful to pay close attention to our units. We want to figure out how much money we made just from adult tickets. We know we sold \(A\) tickets, and each ticket cost $13.50, so we can multiply them together to get our answer: we made \(13.5A\) from adult tickets. Similarly, we made \(6.75C\) from child tickets, so we can finish fillint out our table:
| Adult Tickets | Child Tickets | Total amounts | |
| Tickets Sold | \(A\) | \(C\) | \(41\) |
| Money Made | \({\color{red}13.5A}\) | \({\color{red}6.75C}\) | \({\color{red}398.25}\) |
Step 3: Each row in our table can now be turned into an equation:
Step 4: You can use either substitution or elimination to solve this. We will use substitution for this example. We begin by solving the first equation for \(A\text{:}\)
Now we plug that into the second equation and solve:
We can now plug that into an earlier equation to find \(A\text{:}\)
Step 5: Looking back at our original problem, we wanted to know how many adult and child tickets we sold. This is where choosing good letters for our variables comes in very handy. We have our answer: we sold 18 adult tickets and 23 child tickets.
Checkpoint 9.10.
Suppose you have a store that sells long-sleeve shirts and short-sleeve shirts. Long-sleeve shirts cost $42.19 and short-sleeve shirts cost $32.84. On one day, you sold a total of 80 shirts and made $3029.25. How many of each kind of shirt did you sell?
You sold 43 long sleeve shirts and 37 short-sleeve shirts.
Example 9.11.
In a chemistry experiment, you have two concentrations of hydrochloric acid: 13% solution and 43% solution. How much of each do you need to mix together in order to get 150 mL of solution that is 26.2% concentrated
Step 1:
Weak solution is 13%
Strong solution is 43%
Total liquid: 150 mL
Total acid: 26% of 150mL which is 39.3 mL
Step 2: The two totals we have are total liquid and total acid, so those will be our rows. The two things we don't know are the amount of week solution and amount of strong solution, so those will be our variables. So, our table looks like this:
| Weak Solution | Strong Solution | Total amounts | |
| Total Liquid | Amount of weak solution | Amount of strong solution | Total liquid |
| Total acid | Acid from weak solution | Acid from strong solution | Total acid |
Now, we want to fill in numbers to the inside of the table. Since the things we are looking for are the amount of weak and strong solution, we will make those our variables. We can pick any letters we want, but it's helpful to pick letters that are connected to what they represent. Let's say \(W\) is the amount of weak solution and \(S\) is the amount of strong solution. Therefore, our table looks like this:
| Weak solution | Strong solution | Total amounts | |
| Total liquid | \({\color{red}{W}}\) | \({\color{red}{S}}\) | \({\color{red}150}\) |
| Total acid | Acid from weak solution | Acid from strong solution | Total acid |
Now we just have to fill in the last row, and this is where it will be helpful to pay close attention to our units. We want to figure out how much acid we need from just the weak solution. We know we need \(W\) mL, and it is 13% acid, so we can multiply them together to get our answer: we need \(0.13W\) from the weak solution. Similarly, we need \(0.43S\) from the strong solution, so we can finish filling out our table:
| Weak solution | Strong solution | Total amounts | |
| Total liquid | \(W\) | \(S\) | \(150\) |
| Total acid | \({\color{red}0.13W}\) | \({\color{red}0.43S}\) | \({\color{red}39.3}\) |
Step 3: Each row in our table can now be turned into an equation:
Step 4: You can use either substitution or elimination to solve this. We will use substitution for this example. We begin by solving the first equation for \(W\text{:}\)
Now we plug that into the second equation and solve:
We can now plug that into an earlier equation to find \(W\text{:}\)
Step 5: Looking back at our original problem, we wanted to know how much of each original solution we need. This is where choosing good letters for our variables comes in very handy. We have our answer: we need 84 mL of the weak solution (13% solution) and 66mL of the strong solution (43% solution).
Example 9.15.
You have a total of $500 to invest. You put some of it in an account with Bank A which earns 7% yearly interest. You put the rest in an account with Bank B which earns 13% interest. At the end of the year, you have earned $51.02 in interest. How much did you invest in each account?
Step 1:
Total invested: $500
Bank A earns 7% interest
Bank B earns 13% interest
Total interest earned: $51.02
Step 2: The two totals we have are total money invested and total interest earned, so those will be our rows. The two things we don't know are the amount of money we put in each account, so those will be our variables. So, our table looks like this:
| Bank A | Bank B | Total amounts | |
| Total invested | Amount invested in Bank A | Amount invested in Bank B | Total invested |
| Total interest earned | Interest earned from Bank A | Interest earned from Bank B | Total interest earned |
Now, we want to fill in numbers to the inside of the table. Since the things we are looking for are the amount invested in each account, we will make those our variables. We can pick any letters we want, but it's helpful to pick letters that are connected to what they represent. Let's say \(A\) is the amount invested in Bank A and \(B\) is the amount invested in Bank B. Therefore, our table looks like this:
| Bank A | Bank B | Total amounts | |
| Total invested | \({\color{red}{A}}\) | \({\color{red}{B}}\) | \({\color{red}500}\) |
| Total interest earned | Interest earned from Bank A | Interest earned from Bank B | Total interest earned |
Now we just have to fill in the last row, and this is where it will be helpful to pay close attention to our units. We want to figure out how much interest we earned from just Bank A. We know we put $\(A\) in Bank A and it earns7% interest, so we can multiply them together to get our answer: we earn \(0.07A\) from Bank A. Similarly, we earn \(0.13B\) from Bank B, so we can finish filling out our table:
| Bank A | Bank B | Total amounts | |
| Total invested | \(A\) | \(B\) | \(500\) |
| Total interest earned | \({\color{red}0.07A}\) | \({\color{red}0.13B}\) | \({\color{red}51.02}\) |
Step 3: Each row in our table can now be turned into an equation:
Step 4: You can use either substitution or elimination to solve this. We will use substitution for this example. We begin by solving the first equation for \(A\text{:}\)
Now we plug that into the second equation and solve:
We can now plug that into an earlier equation to find \(W\text{:}\)
Step 5: Looking back at our original problem, we wanted to know how much we invested in each account. This is where choosing good letters for our variables comes in very handy. We have our answer: we put $233 in Bank A and $267 in Bank B.