Solving Systems Of Equations By Substitution A Step-by-Step Guide

Solving systems of equations is a fundamental concept in algebra, and the substitution method is a powerful technique for finding solutions. In this comprehensive guide, we will explore the substitution method in detail, providing a step-by-step approach to solving systems of equations. We will also analyze a specific system to illustrate the application of this method. This article aims to provide a deep understanding of the substitution method, making it accessible to learners of all levels.

Understanding the Substitution Method

The substitution method is an algebraic technique used to solve systems of equations. A system of equations is a set of two or more equations that share the same variables. The goal is to find values for these variables that satisfy all equations simultaneously. The substitution method works by solving one equation for one variable and then substituting that expression into the other equation. This process reduces the system to a single equation with one variable, which can then be easily solved. The value obtained is then substituted back into one of the original equations to find the value of the other variable.

Steps Involved in the Substitution Method

The substitution method involves a series of steps that, when followed correctly, lead to the solution of the system of equations. Let's break down these steps:

1. Solve one equation for one variable: Choose one of the equations in the system and solve it for one of the variables. This means isolating one variable on one side of the equation. Select the equation and variable that appear easiest to isolate. For instance, if one equation has a variable with a coefficient of 1, it might be the easiest choice.
2. Substitute the expression into the other equation: Take the expression obtained in the first step and substitute it into the other equation in the system. This means replacing the selected variable in the second equation with the expression derived from the first equation. The result is a new equation with only one variable.
3. Solve the new equation: Solve the equation obtained in the second step for the remaining variable. This will give you the numerical value of one of the variables in the system.
4. Substitute back to find the other variable: Substitute the value obtained in the third step back into either of the original equations (or the solved equation from step 1) to find the value of the other variable. This process is called back-substitution.
5. Check the solution: Verify that the values obtained for the variables satisfy both original equations. Substitute the values into both equations and ensure that the equations hold true. This step is crucial to ensure the accuracy of the solution.

Advantages of the Substitution Method

The substitution method is particularly useful when one of the equations can be easily solved for one variable. It simplifies the system by reducing it to a single equation, making it easier to solve. This method is versatile and can be applied to various types of systems, including linear and non-linear systems.

Solving a System of Equations Using Substitution An Example

Now, let's apply the substitution method to solve the following system of equations:

$\left\{\begin{array}{c} 2 x+y=6 \\ -8 x-4 y=-24 \end{array} \right.$

Step 1 Solve one equation for one variable

We will solve the first equation, $2x + y = 6$, for $y$. This is the easiest choice because $y$ has a coefficient of 1. To isolate $y$, subtract $2x$ from both sides of the equation:

$y = 6 - 2x$

Step 2 Substitute the expression into the other equation

Next, substitute the expression $6 - 2x$ for $y$ in the second equation, $-8x - 4y = -24$:

$-8x - 4(6 - 2x) = -24$

Step 3 Solve the new equation

Now, solve the equation for $x$. First, distribute the $-4$:

$-8x - 24 + 8x = -24$

Combine like terms:

$0x - 24 = -24$

Add 24 to both sides:

$0 = 0$

Step 4 Interpret the result

The equation $0 = 0$ is a true statement, which means that the system has infinitely many solutions. This occurs when the two equations represent the same line. In this case, the second equation is simply a multiple of the first equation (multiply the first equation by -4 to get the second equation).

Alternative Approach Verification

To further illustrate the concept, we can rewrite the second equation and observe its relationship with the first equation. The second equation is $-8x - 4y = -24$. Divide the entire equation by -4:

$2x + y = 6$

This is the same as the first equation, confirming that the system represents the same line, and thus has infinite solutions.

Identifying Systems with Infinite Solutions or No Solution

Infinite Solutions

As demonstrated in the example above, a system has infinite solutions when the equations are dependent, meaning they represent the same line. This can be identified when, after substitution, the resulting equation is an identity (a true statement regardless of the variable's value), such as $0 = 0$ or $5 = 5$.

No Solution

A system has no solution when the equations are inconsistent, meaning they represent parallel lines that never intersect. This can be identified when, after substitution, the resulting equation is a contradiction (a false statement regardless of the variable's value), such as $0 = 5$ or $-2 = 3$. In such cases, there is no pair of $(x, y)$ values that can satisfy both equations simultaneously.

Applications of Solving Systems of Equations

Solving systems of equations is a crucial skill with numerous applications in mathematics, science, engineering, economics, and everyday life. Understanding how to solve these systems allows us to model and solve real-world problems involving multiple variables and constraints.

Real-World Scenarios

Consider the following real-world scenarios where solving systems of equations can be beneficial:

1. Mixture Problems: Suppose you need to mix two solutions with different concentrations of a substance to obtain a final solution with a desired concentration. This can be modeled using a system of equations.
2. Cost and Revenue Analysis: Businesses often use systems of equations to analyze costs, revenue, and break-even points. For example, determining the number of units a company needs to sell to cover its costs can be solved using a system of equations.
3. Distance, Rate, and Time Problems: Problems involving objects moving at different speeds and directions can be modeled using systems of equations. This is common in physics and engineering.
4. Investment Problems: Financial analysts use systems of equations to model investment portfolios and determine the optimal allocation of assets to achieve specific financial goals.
5. Supply and Demand: In economics, the equilibrium price and quantity of a product can be determined by solving a system of equations representing the supply and demand curves.

Practical Examples

1. Mixing Solutions: A chemist needs to prepare 100 mL of a 30% acid solution by mixing a 10% acid solution and a 50% acid solution. Let $x$ be the amount of the 10% solution and $y$ be the amount of the 50% solution. The system of equations is:

   $\left\{\begin{array}{c} x + y = 100 \\ 0. 10x + 0. 50y = 0. 30(100) \end{array} \right.$

   Solving this system will give the amounts of each solution needed.

2. Cost and Revenue: A company sells two products, A and B. The cost to produce each unit of A is $10, and each unit of B is $15. The selling price for each unit of A is $25, and each unit of B is $35. The company wants to determine the number of units of each product they need to sell to break even. Let $x$ be the number of units of A and $y$ be the number of units of B. The system of equations is:

   $\left\{\begin{array}{c} 25x + 35y = 10x + 15y \\ \end{array} \right.$

   Solving this system will help the company determine the break-even point.

Conclusion Mastering the Substitution Method

In conclusion, the substitution method is a versatile and essential technique for solving systems of equations. By understanding the steps involved and practicing with various examples, you can confidently tackle complex problems and real-world applications. The example provided in this guide demonstrates how to apply the substitution method and interpret the results, including cases with infinite solutions. Mastering this method will significantly enhance your algebraic skills and problem-solving abilities. Remember to always check your solutions to ensure accuracy and a thorough understanding of the process.