Applications of linear systems is a topic that many students find rather challenging and confusing. It part of the grade 10 math course and requires a thorough understanding of how linear relationships in real world work and what a system of two linear equations is. To review what it means to solve a system of two linear equations check out this IntoMath lesson.

In the following blog post we will go over a number of applications of linear equations and provide complete solutions to some common linear systems word problems, such as numbers and mixture problems.

## Numbers and age problems

This is one of the most common applications of linear systems.

**The sum of the two digits of a two digit number is 7. When the digits are reversed, the number is increased by 27. What is the number?**

Obviously, we do not know what the two digits within a number are. Let them be ** x** and

*.*

**y**However, we know that if this is a two digit number, then it contains tens and ones. Therefore, the possible numbers composed of the two digits are:

*(10x + y)* or *(10y + x)*

We know that the sum of the two digits is ** 7**:

**x + y = 7**

and if (10x + y) is the larger number, we get another equation:

*(10x + y) – (10y + x) = 27*

*10x + y – 10y – x = 27*

*9x – 9y = 27*

**x – y = 3**

Now, we can solve the system of the two linear equations to find the values of ** x** and

*, which represent the digits of the number.*

**y**[ 1 ] *x + y = 7*

[ 2 ] *x – y = 3*

[ 1 ] *x = 7 – y*

[ 2 ] *(7 – y) – y = 3*

[ 2 ] *7 – 2y = 3*

[ 2 ] *-2y = -4*

[ 2 ] * y = 2*

[ 1 ] *x = 7 – 2*

[ 1 ] *x = 5*

The two digits of the number are ** 2** and

**. Therefore, the number is**

*5***. If reversed, it is 52 (which is 27 more than 25).**

*25**Twice Ginger’s age increased by 3 times Sue’s age is 103. Five times Ginger’s age decreased by three times Sue’s age is 16. How old will Ginger be five years from now?*

Since we do not know how old neither Ginger nor Sue are, we can express their ages as variables respectively: ** x** and

**y**The first equation is:

*2x + 3y = 103*

The second equation is:

*5x – 3y = 16*

Now, let’s solve the system of the two equations we have created above, in order to get the values of x and y:

[ 1 ] *2x + 3y = 103*

[ 2 ] *5x – 3y = 16*

It makes sense to solve this system using the method of elimination, since there are two identical terms with opposite signs, we can add the two equations:

[ 1 ] *2x + 3y = 103*

[ 2 ] *5x – 3y = 16*

____________________

*7x = 119*

*x = 17*

Since the problem only asks about Ginger’s age, there is no need to solve for Sue’s age. The variable ** x** represents Ginger’s age – 17 years old.

**, Ginger will be**

*In five years**.*

**22 years old**## Mixtures and investment problems

These applications of linear systems are used in Science and Finance.

**Salma has a bottle of 35% salt solution and a bottle of 45% salt solution. How much of each solution should she use to make 500 ml of 43% acetic acid?**

Let the amount in ml of a 35% and a 45% solutions be ** x** and

*respectively. The total amount in ml is 500, therefore:*

**y***x + y = 500*

Now, let’s create a relationship for concentration. Percent should be converted to decimals. There is x ml of a 35% solution – 0.35x and y ml of a 45% solution – 0.45y. The concentration of the total amount is 43% (0.43) times 500 ml = 215:

*0.35x + 0.45y = 215*

Next, let’s set up a system of linear equations:

[ 1 ] *x + y = 500*

[ 2 ] *0.35x + 0.45y = 215*

This system can be solved by subsitution:

[ 1 ] * x = 500 – y*

[ 2 ] *0.35(500 – y) + 0.45y = 215*

[ 2 ] 175 – 0.35y + 0.45y = 215

[ 2 ] *0.1y = 215 – 175*

[ 2 ] *0.1y = 40*

[ 2 ] *y = 400 ml*

[ 1 ] *x = 500 – 400*

[ 1 ] *x =**100 ml*

Therefore, there are 100 ml of 35% solution and 400 ml of 45% solution.

*Sam invested $8000, part at 9% interest rate, and the remainder at 10%interest rate. After one year his total interest from these two investments was $740. How much did he invest at each rate?*

We don’t know how much Sam has invested at each interest rate, so let’s say that at 9% he has invested x amount of dollars and at 10% he has invested y amount of dollars. The total amount in dollars is $8000. Therefore:

*x + y = 8000*

Now we need to create a relationship for interest earned:

*0.09x + 0.1y = 740*

This leads to the system of equations:

[ 1 ] *x + y = 8000*

[ 2 ] *0.09x + 0.1y = 740*

Using the method of substitution, we get:

[ 1 ] *x = 8000 – y*

[ 2 ] *0.09(8000 – y) + 0.1y = 740*

[ 2 ] *720 – 0.09y + 0.1y = 740*

[ 2 ] *0.01y = 20*

[ 2 ] *y = $2000*

[ 1 ] *x = 8000 – 2000*

[ 1 ] *x = $6000*

Therefore, Sam has invested $6000 at 9% interest rate and $2000 at 10% interest rate.

There are multiple real world applications of linear systems. We have gone over only the most common ones above. The key to solving such problems (which are all different) is to understand that in order to create a system of two linear equations there should be two different relationships relating the given two quantities. Once the relationships have been established, the rest simply involves solving the system either by substitution or elimination. The answer should be clearly stated every time and the final statement should always answer the very question asked in the problem.

## There are no comments yet