Solving Systems of Equations in Real-World Contexts

In this lesson we will review various methods for solving systems of linear equations and then learn how to solve systems of nonlinear equations. We will also review methods for solving inequalities and use our knowledge of solving systems of equations and solving inequalities to solve real-world problems. Recall that to solve a system of equations means that we find the values of x and y that make both equations true or that satisfy both equations. In the Pre-Instruction Section of this lesson, we will review the methods of solving systems of linear equations that you learned in the previous lesson.

You learned 3 methods for solving systems of linear equations in the previous lesson:

1. Method of Elimination

2. Method of Substitution

3. Graphing

Method of Elimination
Method of Substitution
Graphing
Self-Check

Example A:

Here is an example of a system of linear equations that we will solve using all three methods. We will give our solution as an ordered pair.

LaTeX: 2x-y=7

LaTeX: x-3y=1

Solution A:

A. By the Method of Elimination

LaTeX: 2x-y=7

LaTeX: x-3y=1

Multiply the 2nd equation by 2 and subtract the new 2nd equation from the first:

LaTeX: 2x-y=7

$LaTeX: 2\left(x-3y\right)=2\left(1\right)$

LaTeX: 2x-y=7

LaTeX: -2x+6y=-2

LaTeX: 5y=5

Solve for y:

LaTeX: y=1

Substitute this value of y into the 2nd equation and solve or x:

$LaTeX: x-3\left(1\right)=1$

LaTeX: x=4

Given as an ordered pair, the solution is (4,1).

Example B:

Here is an example of a system of linear equations that we will solve using all three methods. We will give our solution as an ordered pair.

LaTeX: 2x-y=7

LaTeX: x-3y=1

Solution B:

A. By the Method of Substitution

LaTeX: 2x-y=7

LaTeX: x-3y=1

Solve for x in terms of y in the 2nd equation:

LaTeX: x=1+3y

Substitute this expression in for x into the 1st equation:

$LaTeX: 2\left(1+3y\right)-y=7$

Simplify and solve for y:

LaTeX: 2+6y-y=7

LaTeX: 2+5y=7

LaTeX: 5y=5

LaTeX: y=1

Substitute this value of y into the expression:

LaTeX: x=1+3y

$LaTeX: x=1+3\left(1\right)=4$

Given as an ordered pair, the solution is (4,1)

C. By Graphing

The graph below shows both lines and the intersection point as (4,1)

Screenshot 2024-10-03 at 11.09.00 AM.png

Problem 1:

Solve the following system by Substitution:

LaTeX: 2x+y=10

LaTeX: 3x-y=5

Solution 1:

To solve the system of linear equations by substitution, we will solve one equation for one variable and then substitute that expression into the other equation. Given the equations:

1. LaTeX: 2x+y=10

2. LaTeX: 3x-y=5

Let's solve the second equation for y.

Adding y to both sides, we get:

LaTeX: 3x=y+5

Subtracting 5 from both side, we get:

LaTeX: y=3x-5

Now we will substitute this expression for y into the first equation:

$LaTeX: 2x+\left(3x-5\right)=10$

Now solve for x:

LaTeX: 2x+3x-5=10

LaTeX: 5x-5=10

LaTeX: 5x=10+5

LaTeX: 5x=15

$LaTeX: x=\frac{15}{5}$

LaTeX: x=3

Now that we have x, let's substitute this value into the expression we found for y:

LaTeX: y=3x-5

$LaTeX: y=3\left(3\right)-5$

LaTeX: y=9-5

LaTeX: y=4

The solution to the system of equations is (3,4).

Problem 2:

Solve the following system by elimination

LaTeX: 2x+7=10

LaTeX: 3x-y=5

Solution 2:

To solve the system if linear equations by elimination, we will add the two equations together to eliminate y. Given the equations:

1. LaTeX: 2x+y=10

2. LaTeX: 3x-y=5

Adding these equations together, we get:

LaTeX: 3x=y+5

Subtracting 5 from both side, we get:

$LaTeX: \left(2x+y\right)+\left(3x-y\right)=10+5$

LaTeX: 2x+3x+y-y=15

LaTeX: 5x=15

Now let's solve for x:

LaTeX: 5x=15

$LaTeX: x=\frac{15}{5}$

LaTeX: x=3

Now that we have x, let's substitute this value into one of the original equations to find y. Let's use the first equation:

LaTeX: 2x+y=10

$LaTeX: 2\left(3\right)+y=10$

LaTeX: 6+y=10

LaTeX: y=10-6

LaTeX: y=4

The solution to the system of equations is (3,4).

Problem 3:

Solve the following system by Graphing.

LaTeX: 2x+y=10

LaTeX: 3x-y=5

Solution 3:

Graphing:

Screenshot 2024-10-03 at 1.04.59 PM.png

Instructional Content:

Part I. Solving Systems of Equations Algebraically in Context

In the contextual example from our previous lesson, we will use our knowledge of solving systems of equations to solve the system of equations. Recall the Cutting Board Example from the previous lesson.

Contextual Example: New Mexico Artist’s Cutting Boards
Worked Example
Self-Check

A woodworker in New Mexico makes beautiful art from various types of wood and other natural materials. Below is an example of one of the cutting boards available at Sandia Mountain Woodworks. This particular board is made from 3 types of wood: Walnut, Mahogany and Maple.

Screenshot 2024-09-04 at 11.31.46 PM.png

In this real-world problem, we will explore an example connected to creating cutting boards from only two types of wood: Mahogany and Walnut.

Cutting Board Example

Suppose the artist makes 2 different kinds of cutting boards. Type 1 is made from Walnut and Mahogany. This type of cutting board uses 4 strips of Walnut and 2 strips of Mahogany.

Another type of cutting board, Type 2, also is made from Walnut and Mahogany, but it requires 2 strips of Walnut and 3 strips of Mahogany.

Suppose that the strips of each type of wood used in these two types of cutting boards is the same length and width.

If the total cost of wood material to make Type 1 is approximately $35 and the total cost of wood material to make Type 2 is approximately $30, how much does each strip of each of the types of wood cost?

A system of equations can be used to model this situation.

Let x be the cost of each strip of Mahogany and y be the cost of each strip of Walnut.

Then the system of equations to represent this situation is

LaTeX: 4x+2y=35

LaTeX: 2x+3y=30

Worked Example:

We can use algebraic methods or graphing to solve this system of equations.

Using graphing, you see the solution as the intersection point shown in the graph below:

X = 5.625 and y = 6.25.

Rounding to the nearest cent, this means that each strip of Mahogany costs approximately $5.63 and each strip of Walnut costs $6.25.

Screenshot 2024-10-03 at 11.43.48 AM.png

Self-Check

INSTRUCTIONS: Solve the system of equations from the previous example using the method of Elimination

LaTeX: 4x+2y=45

LaTeX: 2x+3y=40

Round your answers to nearest cent.

Check your answer when ready.

Solution:

To eliminate one of the variables,

we will multiply the second equation by 2 to make the coefficients of x in both equations the same:

LaTeX: 4x+2y=35

LaTeX: 4x+6y=60

Now, we subtract the first equation from the second to eliminate x:

LaTeX: 4x+2y=35

$LaTeX: \left(4x+6y\right)-\left(4x+2y\right)=60-35$

This simplifies to:

LaTeX: 4y=25

Now, let's solve for y:

LaTeX: 4y=25

$LaTeX: y=\frac{25}{4}$

LaTeX: y=6.25

Now substitute $LaTeX: y=\frac{25}{4}$ into one of the original equations to find x. Let's use the first equation:

$LaTeX: 4x+2\left(\frac{25}{4}\right)=35$

$LaTeX: 4x+\frac{25}{2}=35$

$LaTeX: 4x=35-\frac{25}{2}$

$LaTeX: 4x=\frac{70-25}{2}$

$LaTeX: 4x=\frac{45}{2}$

$LaTeX: x=\frac{45}{8}$

So, the solution to the system of equations is $LaTeX: x=\frac{45}{8}$ and $LaTeX: y=\frac{25}{4}$

LaTeX: x=5.625 and LaTeX: y=6.25

Rounded to the nearest cent we get $5.63 for Mahogany and $6.25 for Walnut

Instructional Content:

Part II. Solving System of Equations Containing a Nonlinear Equation

There are systems of equations that involve nonlinear equations. The system shown below is one such example. This system has a linear equation and a quadratic equation.

Worked Example
Self-Check

Worked Example:

We can use the Method of Substitution to solve this system. LaTeX: x^2-y=3,x=1+y Since the 2nd equation has x given in terms of y, we can use this expression for x and substitute it into the 1st equation as follows:

$LaTeX: \left(1+y\right)^2-y=3$

We can expand the binomial and collect like terms to re-write this equation.

LaTeX: 1+2y+y^2-y=3

LaTeX: y^2+y-2=0

This is a quadratic equation. You have learned several methods for solving quadratic equations. Those methods include factoring, using the Quadratic Formula and graphing the equation.

In this case, we can factor this quadratic as follows:

$LaTeX: \left(y+2\right)\left(y-1\right)=0$

Setting each factor this quadratic as follows:

LaTeX: y=-2 or LaTeX: y=1

You still have to solve for x. To do that, you can substitute these values into the 2nd equation

LaTeX: x=1+y

If LaTeX: y=-2 , LaTeX: x=1-2=-1

If LaTeX: y=1 , LaTeX: x=1+1=2

The solutions are (-1,-2) and (2,1).If you graph the two equations in the system, you can see that the graphs intersect in two points.

Screenshot 2024-10-03 at 1.41.19 PM.png

Self-Check

INSTRUCTIONS: Solve the system of equations using the algebraic method of Elimination. Show your steps. Give your solutions as ordered pairs.

LaTeX: 2^2+y=3

LaTeX: x-y=3

Check your answer by using Desmos to graph the two equations and identifying the intersection points.

Solution:

To solve the systems of equations by elimination, we can eliminate one variable by adding or subtracting the equations. Given the equations:

1. LaTeX: x^2+y=3

2. LaTeX: x-y=3

We can eliminate y by adding the two equations together.

Adding equations 1 and equations 2:

$LaTeX: \left(x^2+y\right)+\left(x-y\right)=3+3$

LaTeX: x^2+x=6

Now we have a quadratic equation. Let's rearrange it into standard form:

LaTeX: x^2+x-6=0

Now let's factor the equation and set each factor equal to zero:

$LaTeX: \left(x+2\right)\left(x-3\right)=0$

LaTeX: x-2=0

LaTeX: x+3=0

So the solutions for x are LaTeX: x=2 and LaTeX: x=-3

Now substitute these values of x into one of the original equations to find the corresponding values of y. Let's use equation 2:

For LaTeX: x=2 :

LaTeX: 2-y=3

LaTeX: -y=3-2

LaTeX: -y=1

LaTeX: y=-1

For LaTeX: x=-3 :

LaTeX: -3-y=3

LaTeX: -3-3=y

LaTeX: y=-6

So, the solution to the system of equations are:

1. LaTeX: x=2 , LaTeX: y=-1

2. LaTeX: x=-3 , LaTeX: y=-6

Given the ordered pairs the solutions are (2,-1) and (-3,-6)

Check by graphing:

The graph of the two equations is shown. The two intersection points are identified on the graph.

Instructional Content:

Part III. Using Desmos to Create Artwork

Recall that in TLA 6.1, you learned how to create various figures and shade those figures using inequalities. You may want to refer to the video below to refresh your memory.

Helpful Tools for Desmos Art
Review: YOUTUBE Video
Desmos Example with Simple Figures	Basic Shapes with Shading

Desmos Example
Self-Check

Desmos Example:

You can use your knowledge of graphing inequalities to create fun artwork using Desmos. Here is an example of an adapted version of what some students created using linear and quadratic inequalities and circles.

Screenshot 2024-10-03 at 2.11.41 PM.png

Here is a link to the Desmos file.

You will practice this skill and then use your knowledge and creative imagination to create your own artwork in the Module Assessment.

Self-Check

INSTRUCTIONS: Using the Demos Graphing calculator recreate the image below using linear, quadratic and circle inequalities.

Screenshot 2024-10-03 at 2.14.41 PM.png

Solving Systems of Equations in Real-World Contexts

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