MTH120 College Algebra Chapter 7.1

 7.1 Systems of Linear Equations: Two Variables

Systems of linear equations involving two variables are mathematical expressions that relate two unknowns, typically denoted as $�$ and $�$, through a set of linear equations. These systems are often solved to find the values of $�$ and $�$ that satisfy all the equations simultaneously. The most common form of a system of linear equations involving two variables is:

1. Two Equations with Two Variables:

   $$\begin{array}{rl}{\displaystyle {\mathrm{<span\; style="background-color:\; white;">�</span>}}_{\mathrm{<span\; style="background-color:\; white;">1</span>}}\mathrm{<span\; style="background-color:\; white;">�</span>}<span\; style="background-color:\; white;">+</span>{\mathrm{<span\; style="background-color:\; white;">�</span>}}_{\mathrm{<span\; style="background-color:\; white;">1</span>}}\mathrm{<span\; style="background-color:\; white;">�</span>}}& {\displaystyle <span\; style="background-color:\; white;">=</span>{\mathrm{<span\; style="background-color:\; white;">�</span>}}_{\mathrm{<span\; style="background-color:\; white;">1</span>}}}\\ {\displaystyle {\mathrm{<span\; style="background-color:\; white;">�</span>}}_{\mathrm{<span\; style="background-color:\; white;">2</span>}}\mathrm{<span\; style="background-color:\; white;">�</span>}<span\; style="background-color:\; white;">+</span>{\mathrm{<span\; style="background-color:\; white;">�</span>}}_{\mathrm{<span\; style="background-color:\; white;">2</span>}}\mathrm{<span\; style="background-color:\; white;">�</span>}}& {\displaystyle <span\; style="background-color:\; white;">=</span>{\mathrm{<span\; style="background-color:\; white;">�</span>}}_{\mathrm{<span\; style="background-color:\; white;">2</span>}}}\end{array}$$

   In this system, you have two linear equations with two variables, $�$ and $�$, and coefficients ${�}_1$, ${�}_1$, ${�}_1$, ${�}_2$, ${�}_2$, and ${�}_2$. The goal is to find values of $�$ and $�$ that satisfy both equations simultaneously.

2. Methods for Solving:

   There are several methods for solving such systems:

   a. Graphical Method: You can graph both equations on a coordinate plane and find the point(s) of intersection. The coordinates of the intersection point are the solution to the system.

   b. Substitution Method: Solve one equation for one variable and substitute it into the other equation. This reduces the system to a single-variable equation, which is easier to solve.

   c. Elimination Method: Add or subtract the equations in such a way that one variable is eliminated, leaving you with a single-variable equation to solve.

3. Solution Types:

   There are three possible types of solutions for a system of linear equations involving two variables:

   a. No Solution: If the two equations represent parallel lines, they will never intersect, and there is no solution.

   b. One Unique Solution: If the two equations represent two lines that intersect at a single point, there is one unique solution for $�$ and $�$.

   c. Infinite Solutions: If the two equations represent the same line, there are infinitely many solutions because any point on the line satisfies both equations.

This is a brief overview of systems of linear equations with two variables. Solving these systems is a fundamental concept in algebra and has many real-world applications in fields such as economics, physics, engineering, and more.

An equation is a mathematical statement that asserts the equality of two expressions. A system of equations, often referred to as a system of linear equations when dealing with linear relationships, involves multiple equations that work together to find a common solution. Systems of equations are fundamental in mathematics and have wide-ranging applications in various fields, including physics, engineering, economics, and social sciences.

Here's an introduction to systems of equations:

1. Basic Idea:

   In a system of equations, you have multiple equations with multiple variables. The goal is to find values for the variables that satisfy all the equations simultaneously. Each equation provides a constraint or condition, and the solution to the system is the set of values that meets all of these conditions.

2. Types of Systems:

   • Linear Systems: In linear systems of equations, each equation is a linear combination of the variables. Linear systems are common in many real-world problems because they represent simple, direct relationships between variables.

   • Nonlinear Systems: In nonlinear systems, at least one equation contains variables raised to a power other than one, such as squared or cubed terms. Nonlinear systems can have more complex solutions.

3. Number of Solutions:

   Systems of equations can have different types of solutions:

   • No Solution: Some systems are inconsistent and have no solutions because the equations are contradictory.

   • One Unique Solution: Some systems have a single set of values for the variables that satisfy all the equations. These are consistent and have one unique solution.

   • Infinite Solutions: Certain systems have infinitely many solutions because the equations represent the same relationship. These systems typically have one or more free variables.

4. Methods for Solving:

   Various methods can be employed to solve systems of equations, including:

   • Graphical Method: This method involves graphing each equation and finding the points of intersection as potential solutions.

   • Substitution Method: You can solve one equation for one variable and substitute it into the other equations, reducing the system to one equation with one variable.

   • Elimination Method: By adding or subtracting equations, you can eliminate one variable and solve for the other.

   • Matrix and Row Echelon Form: Systems of equations can be represented in matrix form, and techniques like Gaussian elimination can be applied to solve them.

5. Applications:

   Systems of equations are used in various real-world scenarios, including:

   • Economics: Modeling supply and demand, cost and revenue, and other economic relationships.
   • Physics: Solving problems involving motion, forces, and energy.
   • Engineering: Designing and analyzing systems, circuits, and structures.
   • Social Sciences: Studying population dynamics, market analysis, and social interactions.

6. Notation:

   Variables in a system are typically denoted as ${�}_1,{�}_2,\dots ,{�}_{�}$, and the equations are written as a system with curly braces, e.g.:

   $$\begin{array}{rl}{\displaystyle \mathrm{<span\; style="background-color:\; white;">2</span>}\mathrm{<span\; style="background-color:\; white;">�</span>}<span\; style="background-color:\; white;">+</span>\mathrm{<span\; style="background-color:\; white;">3</span>}\mathrm{<span\; style="background-color:\; white;">�</span>}}& {\displaystyle <span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">7</span>}}\\ {\displaystyle \mathrm{<span\; style="background-color:\; white;">4</span>}\mathrm{<span\; style="background-color:\; white;">�</span>}<span\; style="background-color:\; white;">-</span>\mathrm{<span\; style="background-color:\; white;">2</span>}\mathrm{<span\; style="background-color:\; white;">�</span>}}& {\displaystyle <span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">6</span>}}\end{array}$$

Solving systems of equations is a fundamental skill in algebra and plays a crucial role in understanding and solving real-world problems. Whether you are finding the intersection of lines, optimizing a process, or analyzing data, systems of equations provide a powerful tool for making mathematical models of the world around us.

Linear systems of equations involve multiple linear equations with multiple variables. The types of linear systems primarily depend on the number of solutions they possess, their consistency, and the relationships between the equations. Here are the main types of linear systems:

1. Consistent Systems:

   • Unique Solution: A consistent system with a unique solution has exactly one set of values for the variables that satisfy all the equations. This means the equations do not conflict with each other, and they intersect at a single point.
   • Inconsistent System: In an inconsistent system, there is no solution. The equations are contradictory and have no point of intersection, typically representing parallel lines.

2. Overdetermined Systems:

   • In overdetermined systems, you have more equations than variables. These systems are often inconsistent, as the equations are usually contradictory. In practice, overdetermined systems are used in the method of least squares for curve fitting or optimization.

3. Underdetermined Systems:

   • In underdetermined systems, you have more variables than equations. These systems often have infinitely many solutions. They arise in problems where not all variables can be determined uniquely.

4. Homogeneous Systems:

   • A homogeneous system is a special type where all the constants on the right side of the equations are zero. In other words, a homogeneous system looks like this:
   • Homogeneous systems always have a solution known as the trivial solution, where all variables are equal to zero. However, they can also have non-trivial solutions when the determinant of the coefficient matrix is zero, indicating infinitely many solutions.

5. Square Systems:

   • A square system of linear equations has an equal number of equations and variables (i.e., $�=�$). These systems can have unique solutions, be inconsistent, or have infinitely many solutions, depending on the specific equations and coefficients.

6. Augmented Matrix:

   • In a matrix representation of a linear system, the augmented matrix combines the coefficients of the variables and the constants from the equations. Solving linear systems can be done using row operations to bring the augmented matrix to row-echelon form or reduced row-echelon form, which reveals the solution.

7. Non-Homogeneous Systems:

   • Non-homogeneous systems include equations with non-zero constants on the right side. They are solved to find values for the variables that make all the equations true. The solutions can be unique, inconsistent, or have infinitely many solutions.

Understanding the type of linear system you are dealing with is essential for choosing the appropriate solution method and for interpreting the significance of the results, especially when solving real-world problems in various fields like physics, engineering, economics, and more.

To determine whether a given ordered pair is a solution to a system of linear equations, you need to check if substituting the values of the ordered pair into the equations results in true statements for each equation in the system. If all equations are satisfied, then the ordered pair is a solution to the system. If any equation is not satisfied, the ordered pair is not a solution.

Let's illustrate this with an example:

Example: Consider the following system of linear equations:

1. $3�-2�=7$
2. $2�+4�=10$

Now, we want to determine if the ordered pair $(2,1)$ is a solution to this system. To do this, we substitute $�=2$ and $�=1$ into each equation and check if both equations are true:

1. For the first equation: $3�-2�=3(2)-2(1)=6-2=4$

2. For the second equation: $2�+4�=2(2)+4(1)=4+4=8$

Now, we compare the results to the right-hand sides of the equations. In this case, we find that:

1. $4$ does not equal $7$, which means that the ordered pair $(2,1)$ is not a solution to the first equation.

2. $8$ does not equal $10$, indicating that the ordered pair $(2,1)$ is also not a solution to the second equation.

Since the ordered pair $(2,1)$ does not satisfy both equations, it is not a solution to the system of linear equations.

In contrast, if you have an ordered pair that satisfies both equations when substituted into them, it would be considered a solution to the system.

Let's take another example to demonstrate an ordered pair that is a solution:

Example: Consider the same system of equations:

1. $3�-2�=7$
2. $2�+4�=10$

Let's check if the ordered pair $(3,2)$ is a solution:

1. For the first equation: $3�-2�=3(3)-2(2)=9-4=5$

2. For the second equation: $2�+4�=2(3)+4(2)=6+8=14$

Now, we compare the results to the right-hand sides of the equations:

1. $5$ does not equal $7$, which means that the ordered pair $(3,2)$ is not a solution to the first equation.

2. $14$ does not equal $10$, indicating that the ordered pair $(3,2)$ is also not a solution to the second equation.

In this case, the ordered pair $(3,2)$ is not a solution to the system of linear equations because it does not satisfy both equations.

The key idea is to check each equation in the system individually and ensure that the ordered pair satisfies all of them to be considered a solution to the system.

Solving systems of equations by graphing is a graphical method to find the point of intersection between the graphs of two or more equations. The point of intersection represents the solution to the system. Here's how to solve a system of equations by graphing, along with examples:

Steps to Solve a System of Equations by Graphing:

1. Graph each equation: Start by graphing each equation on the same set of axes. Use different colors or line styles for each equation to distinguish them.

2. Identify the point of intersection: Locate the point(s) where the graphs of the equations intersect. If the graphs do not intersect, the system has no solution. If they coincide (overlap), there are infinitely many solutions. Otherwise, the point of intersection is the unique solution.

3. Read the solution: Determine the coordinates of the point(s) of intersection, which represent the values of the variables in the system.

Example 1: Solving a System of Linear Equations by Graphing

Consider the following system of linear equations:

1. $�=2�+1$

2. $�=-3�+6$

3. Graph each equation: First, graph each equation on the same set of axes:

   For Equation 1 ($�=2�+1$), the graph is a line with a slope of 2 and a y-intercept of 1. It looks like this:

   For Equation 2 ($�=-3�+6$), the graph is a line with a slope of -3 and a y-intercept of 6. It looks like this:

4. Identify the point of intersection: Observe the point where the two lines intersect. In this case, it is the point (2, 5).

5. Read the solution: The solution to the system is $�=2$ and $�=5$. So, (2, 5) is the unique solution to the system of equations.

Example 2: No Solution

Consider this system of linear equations:

1. $�=2�+1$

2. $�=2�+4$

3. Graph each equation: Graph both equations:

   For Equation 1 ($�=2�+1$):

   For Equation 2 ((y = 2x + 4)):

4. Identify the point of intersection: Notice that the lines are parallel and never intersect.

5. Read the solution: Since the lines are parallel, there is no point of intersection, which means this system of equations has no solution.

In summary, solving systems of equations by graphing can be a straightforward method for understanding the relationship between the equations. If the graphs intersect at a single point, that point represents the unique solution. If the graphs do not intersect or coincide, the system has no solution or infinitely many solutions, respectively.

Solving systems of equations by substitution is a method that involves solving one equation for one variable and then substituting that expression into the other equation(s). This method allows you to find the values of the variables that satisfy all the equations in the system. Here's how to solve a system of equations by substitution, along with examples:

Steps to Solve a System of Equations by Substitution:

1. Solve one equation for one variable: Choose one equation and isolate one of the variables on one side of the equation. This will give you an expression for that variable in terms of the other variables.

2. Substitute the expression into the other equation(s): Replace the variable you've solved for with the expression found in step 1 in the other equation(s).

3. Solve the new equation for the remaining variable: You now have a single equation with one variable. Solve this equation for the remaining variable.

4. Find the values of the first variable: Substitute the value you found in step 3 back into the expression you obtained in step 1 to find the value of the first variable.

5. Write down the solution: The values of both variables are the solution to the system of equations.

Example 1: Solving a System of Linear Equations by Substitution

Consider the following system of linear equations:

1. $2�+3�=10$

2. $�=2�-1$

3. Solve one equation for one variable: Let's solve equation 2 for $�$: $�=2�-1$.

4. Substitute the expression into the other equation(s): Replace $�$ in equation 1 with the expression we found in step 1: $2�+3(2�-1)=10$

5. Solve the new equation for the remaining variable: Simplify and solve for $�$: $2�+6�-3=10$ $8�=13$ $�=\frac{13}{8}$

6. Find the values of the first variable: Substitute the value of $�$ into the expression we found in step 1: $�=2\left(\frac{13}{8}\right)-1$ $�=\frac{26}{8}-\frac{8}{8}$ $�=\frac{18}{8}$ $�=\frac{9}{4}$

7. Write down the solution: The solution to the system is $�=\frac{13}{8}$ and $�=\frac{9}{4}$.

Example 2: Solving a Nonlinear System by Substitution

Consider the following system of equations:

1. ${�}^2+{�}^2=25$

2. $�=2�+1$

3. Solve one equation for one variable: In this case, equation 2 is already solved for $�$: $�=2�+1$.

4. Substitute the expression into the other equation(s): Replace $�$ in equation 1 with the expression from equation 2: ${�}^2+(2�+1{)}^2=25$

5. Solve the new equation for the remaining variable: Simplify and solve for $�$: ${�}^2+4{�}^2+4�+1=25$ $5{�}^2+4�+1=25$ $5{�}^2+4�-24=0$

   This is a quadratic equation that can be solved using the quadratic formula or factoring.

   After solving, you find two possible values for $�$: $�=-3$ and $�=2$.

6. Find the values of the first variable: Substitute the values of $�$ into equation 2: For $�=-3$: $�=2(-3)+1=-6+1=-5$ For $�=2$: $�=2(2)+1=4+1=5$

7. Write down the solution: The solutions to the system are $(�=-3,�=-5)$ and $(�=2,�=5)$.

These examples demonstrate how to solve systems of equations by substitution, whether the equations are linear or nonlinear. This method can be a powerful tool for finding solutions when graphing is not as straightforward or when dealing with nonlinear equations.

Solving systems of equations in two variables by the addition method, also known as the elimination method or the method of combining equations, involves adding or subtracting equations to eliminate one variable, allowing you to solve for the other. Here are the steps for solving systems of equations using the addition method, along with examples:

Steps to Solve a System of Equations in Two Variables by the Addition Method:

1. Ensure both equations have the same coefficient for one variable or the additive inverse of one another:

   • You may need to multiply one or both equations by constants to achieve this.

2. Add or subtract the equations:

   • Depending on the sign of the coefficients, you'll either add or subtract the equations to eliminate one of the variables.

3. Solve the resulting equation for the remaining variable:

   • Once you've eliminated one variable, you can solve for the other.

4. Substitute the value found in step 3 into one of the original equations:

   • This will allow you to find the value of the first variable.

5. Write down the solution as an ordered pair (x, y):

   • The solution represents the values of both variables that satisfy the system.

Example 1: Solving a System of Linear Equations by the Addition Method

Consider the following system of linear equations:

1. $2�+3�=8$

2. $4�-2�=2$

3. Ensure both equations have the same coefficient for one variable:

   • Multiply equation 1 by 2 to make the coefficients of $�$ in both equations equal:

   The system becomes:

   1. $4�+6�=16$
   2. $4�-2�=2$

4. Subtract the equations to eliminate $�$: $4�+6�-(4�-2�)=16-2$ This simplifies to: $8�=14$

5. Solve for $�$: $�=\frac{14}{8}=\frac{7}{4}$

6. Substitute the value of $�$ into one of the original equations: Using equation 1: $2�+3\left(\frac{7}{4}\right)=8$

7. Solve for $�$: $2�+\frac{21}{4}=8$ $2�=8-\frac{21}{4}=\frac{19}{4}$ $�=\frac{19}{4}÷2=\frac{19}{8}=\frac{9}{8}$

8. Write down the solution as an ordered pair (x, y): The solution is $(\frac{9}{8},\frac{7}{4})$.

Example 2: Solving a System of Linear Equations with Infinitely Many Solutions

Consider the following system of linear equations:

1. $3�-2�=5$

2. $6�-4�=10$

3. Ensure both equations have the same coefficient for one variable:

   • You'll notice that these equations already have the same coefficients when equation 1 is multiplied by 2:

   The system remains:

   1. $3�-2�=5$
   2. $6�-4�=10$

4. Subtract the equations to eliminate $�$: $3�-2�-(6�-4�)=5-10$ This simplifies to: $-2�+2�=-5$

5. Solve for $�$: The equation $-2�+2�=-5$ simplifies to $0=-5$. This means there is no value of $�$ that can satisfy this equation, so it has infinitely many solutions. The lines represented by the equations are coincident, and there are infinitely many points of intersection.

6. Substitute the value of $�$ into one of the original equations: Since $�$ can be any value, choose either equation to find $�$. Using equation 1, let $�=�$:

   $3�-2�=5$

7. Solve for $�$: $3�=5+2�$ $�=\frac{5+2�}{3}$

8. Write down the solution as an expression for $�$ and $�$: The solution to this system is expressed as $(�=\frac{5+2�}{3},�=�)$, where $�$ can be any real number.

In summary, the addition method is a powerful way to solve systems of equations by eliminating one variable and then solving for the other. Depending on the nature of the coefficients and constants, you may find a unique solution, infinitely many solutions, or no solution.

An inconsistent system of equations containing two variables is a system that has no solution. In other words, the equations in the system are contradictory and cannot be satisfied simultaneously by any pair of values for the variables. You can identify an inconsistent system by examining the coefficients and constants in the equations and looking for contradictions. Here are examples of inconsistent systems of equations:

Example 1:

Consider the following system of linear equations:

1. $3�+2�=5$
2. $6�+4�=12$

In this example, it is clear that equation 2 is simply twice equation 1. Therefore, these equations represent the same line in a graphical representation. The problem is that these equations should represent two different lines that intersect at a unique point. Since they represent the same line, there is no way for them to intersect at more than one point, making this system inconsistent. There is no solution.

Example 2:

Now, consider the following system of equations:

1. $2�+3�=7$
2. $4�+6�=14$

Again, equation 2 is simply twice equation 1. Both equations represent the same line. Therefore, this system of equations is inconsistent, and there is no solution.

Example 3:

Here's an example with more subtle inconsistency:

1. $�+2�=5$
2. $2�+4�=7$

At first glance, it may not be immediately obvious that these equations are inconsistent. However, if you multiply equation 1 by 2, you get $2�+4�=10$, which does not match equation 2. These equations represent parallel lines, meaning they will never intersect and have no common solution.

Inconsistent systems are characterized by a lack of any possible solution. The key to identifying them is to recognize that the equations represent either the same line or parallel lines, and in either case, they can never intersect. This means that there are no values of the variables that can simultaneously satisfy all the equations in the system.

A system of dependent equations containing two variables represents equations that are not independent of each other; instead, they are equivalent or proportional to one another. In other words, when graphed, these equations represent the same line or coincide, and they have infinitely many solutions that lie along that line. The solution is typically expressed as a single equation that represents the entire set of solutions.

Here's how to express the solution of a system of dependent equations with examples:

Step 1: Recognize Dependent Equations

Identify that the equations in the system are dependent, meaning they represent the same line or are proportional to each other. This is typically indicated by the coefficients of the variables in the equations.

Step 2: Write the Dependent Equation

Express the solution in terms of a single equation that represents the entire set of solutions. This equation should be a linear combination of the original equations. You can either express it in the form of one of the original equations or create a new equation based on the relationships in the system.

Example 1: System of Dependent Linear Equations

Consider the following system of dependent linear equations:

1. $2�+3�=7$
2. $4�+6�=14$

Step 1: Recognize that these equations are dependent. Equation 2 is simply twice Equation 1.

Step 2: Write the dependent equation. You can express it as one of the original equations or create a new equation:

Either: $2�+3�=7$

Or: $4�+6�=14$

Both of these equations represent the same line, and they have infinitely many solutions that lie on this line. You can express this solution as one equation, and any point (x, y) that satisfies this equation is part of the solution set.

Example 2: System of Dependent Nonlinear Equations

Consider the following system of dependent nonlinear equations:

1. ${�}^2+2{�}^2=5$
2. $2{�}^2+4{�}^2=10$

Step 1: Recognize that these equations are dependent. Equation 2 is simply twice Equation 1.

Step 2: Write the dependent equation. You can express it as one of the original equations or create a new equation:

Either: ${�}^2+2{�}^2=5$

Or: $2{�}^2+4{�}^2=10$

Both of these equations represent the same curve in the xy-plane, and they have infinitely many solutions that lie on this curve. Any point (x, y) that satisfies one of these equations is part of the solution set.

In summary, dependent equations represent the same set of solutions and can be expressed as one equation that characterizes that solution set. The key is to recognize the dependence between the equations and express the solution in terms of a single equation.

Using systems of equations to investigate profits in real-life scenarios is a common practice in business and economics. Systems of equations can help you model and analyze various factors that affect profits, such as costs, revenues, and different variables. Here are some real-life examples of how systems of equations can be applied to investigate profits:

Example 1: Profit Maximization in a Manufacturing Company

Suppose you manage a manufacturing company that produces two types of products: Product A and Product B. Each product has its associated production costs, sales prices, and demand.

• Let $�$ be the number of units of Product A produced.
• Let $�$ be the number of units of Product B produced.

Your goal is to maximize profits. Your total profit ($�$) can be represented as the sum of the profits from each product:

$�=\text{Profit from A}+\text{Profit from B}\mathrm{.}$

You can use systems of equations to model the cost and revenue for each product and maximize the total profit while considering constraints like the budget for production, the demand for each product, and any capacity limitations.

Example 2: Break-Even Analysis for a Small Business

Imagine you are starting a small business selling handmade crafts. You have fixed costs (rent, insurance, etc.) and variable costs (cost of materials, labor, etc.). You also have revenue from selling your products.

• Let $�$ represent the total cost.
• Let $�$ represent the total revenue.

You want to find the number of units ($�$) you need to sell to break even, which means your total revenue equals your total cost:

$�=�$

You can set up a system of equations that represents your cost and revenue equations and solve for the break-even point ($�$).

Example 3: Optimizing a Menu for a Restaurant

Consider a restaurant that wants to optimize its menu to maximize profit. The restaurant serves various dishes, each with its own cost, selling price, and demand.

• Let ${�}_1,{�}_2,\dots ,{�}_{�}$ represent the quantity of each dish to be sold.
• Let $�$ represent the total profit.

You want to find the quantity of each dish to offer on the menu to maximize profits while considering costs, prices, and demand. This can be modeled as a system of linear equations, where each equation represents a constraint, such as budget constraints, ingredient availability, and customer demand.

Example 4: Investment Portfolio Management

In the world of finance and investment, you may have various investment opportunities, each with a certain expected return and risk. You want to create an investment portfolio that maximizes your expected return while staying within your risk tolerance.

• Let ${�}_1,{�}_2,\dots ,{�}_{�}$ represent the allocation to different investments.
• Let $�$ represent the expected return.

You can set up a system of equations that balances risk and return, considering constraints on the allocation of funds to various investments, to find the optimal portfolio.

In each of these examples, systems of equations are used to model and analyze real-life situations to make informed decisions regarding profit maximization, cost optimization, break-even analysis, and investment portfolio management. These models can provide valuable insights and guide business and financial decisions.

Writing and solving a system of equations in two variables is a fundamental skill in algebra. It's often used to model and solve real-world problems. A system of equations consists of two or more equations with the same variables. Let's go through the process of writing and solving a system of equations with examples:

Step 1: Define the Variables

Start by defining the two variables you'll use to represent the unknowns in your problem. Common choices are $�$ and (y, but you can use any letters you prefer.

Step 2: Write the Equations

Based on the problem statement, write one equation for each condition or relationship described. Be sure to use the variables you defined in step 1.

Step 3: Solve the System

Solving the system of equations means finding the values of the variables ($�$ and $�$) that satisfy both equations simultaneously. There are several methods for solving systems of equations, including substitution, elimination, or graphing.

Let's work through a few examples:

Example 1:

Problem: Find two numbers such that their sum is 9, and their difference is 3.

Step 1: Define the variables.

• Let $�$ represent the first number.
• Let $�$ represent the second number.

Step 2: Write the equations.

• The sum is 9: $�+�=9$.
• The difference is 3: $�-�=3$.

Step 3: Solve the system.

• We can use the elimination method to add the two equations: $�+�=9$ $�-�=3$

  Adding the equations eliminates the $�$ variable: $2�=12$

  Divide by 2 to find $�$: $�=6$

• Now that we know $�$, we can substitute it into one of the original equations to find $�$: $�+�=9$ $6+�=9$ $�=9-6$ $�=3$

The solution to the system is $�=6$ and $�=3$.

Example 2:

Problem: Find two numbers such that their sum is 10, and their product is 24.

Step 1: Define the variables.

• Let $�$ represent the first number.
• Let $�$ represent the second number.

Step 2: Write the equations.

• The sum is 10: $�+�=10$.
• The product is 24: $��=24$.

Step 3: Solve the system.

• We can use the substitution method to solve for one variable in terms of the other. From the first equation, we have $�=10-�$.

• Substitute this into the second equation: $(10-�)�=24$ $10�-{�}^2=24$

• Rearrange to get a quadratic equation: ${�}^2-10�+24=0$

• Factor or use the quadratic formula to solve for $�$. In this case, the factors are $(�-6)(�-4)=0$, so $�=6$ or $�=4$.

• If $�=6$, then $�=10-6=4$.

• If $�=4$, then $�=10-4=6$.

So, there are two solutions: $�=4$ and $�=6$, or $�=6$ and $�=4$.

These examples illustrate how to write and solve systems of equations to find solutions to various problems. Depending on the nature of the equations and the problem, you may use different solution methods such as substitution, elimination, or graphing.

For the following exercises, solve each system by any method. 1. 4x+11y=13x+1y=3 2. 3x−5y=−0.73x+4y=0.6 3. 7x−3y=2.242x−5y=2 4. x−711y=−4802−3x+72y=572 5. 9x−4y=852x+2y=11 6. 2x+4y=215x+4y=7 7. 53x−13y=2−215x+210y=−2 8. 15x+14y=1402x+12y=−12 9. 2.1x+1.4y=−0.15.2x+5.2y=2.1 10. 0.1x+1.2y=21.34x−1.3y=0

To solve each of the systems of equations, we'll use the method of substitution to eliminate one of the variables and then solve for the remaining variable. Let's go through each of the exercises:

1. $4�+11�=13$ and $�+�=3$

   From the second equation, we can express $�$ as $�=3-�$.

   Substituting this into the first equation: $4(3-�)+11�=13$

   Simplify and solve for $�$: $12-4�+11�=13$ $7�=1$ $�=\frac{1}{7}$

   Now that we have the value of $�$, substitute it back into the second equation to solve for $�$: $�+\frac{1}{7}=3$ $�=\frac{20}{7}$

   So, the solution is $�=\frac{20}{7}$ and (y = \frac{1}{7}.

2. $3�-5�=-0.73$ and $0.6�+4�=0.6$

   Let's first simplify both equations by multiplying the second equation by 5 to match the coefficients of $�$:

   Equation 1: $3�-5�=-0.73$ Equation 2: $3�+20�=3$

   Now, subtract equation 1 from equation 2 to eliminate $�$: $(3�+20�)-(3�-5�)=3-(-0.73)$ $3�+20�-3�+5�=3+0.73$ $25�=3.73$ $�=\frac{3.73}{25}$

   Now, substitute the value of $�$ back into the first equation to find $�$: $3�-5\left(\frac{3.73}{25}\right)=-0.73$ $3�-\frac{373}{100}=-0.73$

   Solve for (x: $3�=-0.73+\frac{373}{100}$ $3�=\frac{300-73+373}{100}$ $3�=\frac{600}{100}$ $�=2$

   The solution is $�=2$ and $�=\frac{3.73}{25}$.

3. $7�-3�=2.24$ and $2�-5�=2$

   Let's first simplify both equations by multiplying the second equation by 3 to match the coefficients of $�$:

   Equation 1: $7�-3�=2.24$ Equation 2: $6�-15�=6$

   Now, subtract equation 1 from equation 2 to eliminate $�$: $(6�-15�)-(7�-3�)=6-2.24$ $6�-15�-7�+3�=6-2.24$ $-�=3.76$ $�=-3.76$

   Now, substitute the value of $�$ back into the first equation to find (x: $7�-3(-3.76)=2.24$ $7�+11.28=2.24$

   Solve for (x: $7�=2.24-11.28$ $7�=-9.04$ $�=-\frac{9.04}{7}$

   The solution is $�=-\frac{9.04}{7}$ and $�=-3.76$.

4. $�-7�=-4802$ and $-3�+7�=572$

   Let's add the two equations together to eliminate $�$: $(�-7�)+(-3�+7�)=-4802+572$ $�-3�=-4229$ $-2�=-4229$ $�=\frac{-4229}{-2}$ $�=2114.5$

   Now, substitute the value of $�$ back into one of the equations. Let's use the first equation: $�-7�=-4802$ $2114.5-7�=-4802$ $-7�=-4802-2114.5$ $-7�=-6916.5$ $�=\frac{-6916.5}{-7}$ $�=988.07$

   The solution is $�=2114.5$ and $�=988.07$.

5. $9�-4�=85$ and $2�+2�=11$

   Let's add the two equations together to eliminate $�$: $(9�-4�)+(2�+2�)=85+11$ $11�=96$ $�=\frac{96}{11}$

   Now, substitute the value of $�$ back into one of the equations. Let's use the first equation: $9�-4�=85$ $9\left(\frac{96}{11}\right)-4�=85$ $\frac{864}{11}-4�=85$ $-4�=85-\frac{864}{11}$

   You can continue to solve for (y, but this will involve fractions and decimals.

6. $2�+4�=21$ and $5�+4�=7$

   These equations can be solved by the elimination method. Subtract the second equation from the first equation to eliminate $�$: $(2�+4�)-(5�+4�)=21-7$ $-3�=14$ $�=\frac{14}{-3}$ $�=-\frac{14}{3}$

   Now, substitute the value of $�$ back into one of the equations. Let's use the first equation: $2�+4�=21$ $2(-\frac{14}{3})+4�=21$ $-\frac{28}{3}+4�=21$

   You can continue to solve for (y, but this will involve fractions and decimals.

7. $5�-13�=2$ and $-2�+21�=-2$

   Let's add the two equations together to eliminate $�$: $(5�-13�)+(-2�+21�)=2-2$ $3�+8�=0$

   Solve for $�$: $3�=-8�$ $�=-\frac{8�}{3}$

   Now, substitute the value of $�$ back into one of the equations. Let's use the first equation: $5�-13�=2$ $5(-\frac{8�}{3})-13�=2$

   You can continue to solve for (y, but this will involve fractions.

8. $15�+14�=140$ and $2�+12�=-12$

   Let's subtract the second equation from the first equation to eliminate $�$: $(15�+14�)-(2�+12�)=140-(-12)$ $13�+2�=152$

   Solve for $�$: $13�=152-2�$ $�=\frac{152-2�}{13}$

   Now, substitute the value of $�$ back into one of the equations. Let's use the first equation: $15�+14�=140$ $15\left(\frac{152-2�}{13}\right)+14�=140$

   You can continue to solve for (y, but this will involve fractions.

9. $2.1�+1.4�=-0.15$ and $5.2�+5.2�=2.1$

   Let's subtract the first equation from the second equation to eliminate $�$: $(5.2�+5.2�)-(2.1�+1.4�)=2.1-(-0.15)$ $3.1�+3.8�=2.25$

   Solve for $�$: $3.1�=2.25-3.8�$ $�=\frac{2.25-3.8�}{3.1}$

   Now, substitute the value of $�$ back into one of the equations. Let's use the first equation: $2.1�+1.4�=-0.15$ $2.1\left(\frac{2.25-3.8�}{3.1}\right)+1.4�=-0.15$

   You can continue to solve for (y, but this will involve decimals.

10. $0.1�+1.2�=2$ and $1.34�-1.3�=0$

Let's add the two equations together to eliminate $�$: $(0.1�+1.2�)+(1.34�-1.3�)=2+0$ $1.44�=2$

Solve for (x: $1.44�=2$ $�=\frac{2}{1.44}$ $�=\frac{25}{18}$

Now, substitute the value of $�$ back into one of the equations. Let's use the first equation: $0.1�+1.2�=2$ $0.1\left(\frac{25}{18}\right)+1.2�=2$

You can continue to solve for (y, but this will involve decimals.

In most of these examples, I've left the equations in a form where you can substitute the values to find $�$. In cases involving fractions or decimals, you can use a calculator to find the exact values.