Solving Systems Of Equations By Substitution Method

by ADMIN 52 views

Solving systems of equations is a fundamental concept in algebra, and the substitution method is a powerful tool for finding solutions. This article will delve into how to use the substitution method to solve a system of equations, with a particular focus on systems involving both quadratic and linear equations. We'll also emphasize the importance of checking your solutions to ensure accuracy. Let's consider the system:

$\left\{\begin{aligned} x^2-4 x+y^2 & =-3 \\ x-y & =3 \end{aligned}\right.$

This article will guide you through the step-by-step process of solving this system using substitution and verifying the solutions. Understanding the nuances of this method is crucial for various applications in mathematics, science, and engineering. We will break down each step, providing clear explanations and insights to enhance your problem-solving skills.

Understanding the Substitution Method

The substitution method involves 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 a single variable, which can then be solved using standard algebraic techniques. This method is particularly effective when one of the equations can be easily solved for one variable in terms of the other. In our case, the second equation, x - y = 3, is a perfect candidate for this. By isolating one variable, we can create an expression to substitute into the first equation, a quadratic equation, and simplify the system significantly.

Step-by-Step Breakdown of the Substitution Method

To effectively use the substitution method, it's crucial to follow a structured approach. The method typically involves these steps:

  1. Isolate one variable in one of the equations: Choose the equation and variable that are easiest to isolate. This often means selecting an equation where the variable has a coefficient of 1. This simplifies the expression and reduces the chances of errors in subsequent steps.
  2. Substitute the expression into the other equation: Replace the isolated variable in the second equation with the expression obtained in the first step. This will result in an equation with only one variable.
  3. Solve the resulting equation: Solve the new equation for the remaining variable. This might involve using algebraic techniques such as factoring, the quadratic formula, or other methods depending on the equation's complexity. The goal is to find all possible values for this variable.
  4. Substitute back to find the other variable: Once you have the value(s) of one variable, substitute them back into either of the original equations (or the isolated expression from step 1) to find the corresponding value(s) of the other variable. This step is crucial for completing the solution and finding all pairs of values that satisfy both equations.
  5. Check your solutions: Always verify your solutions by substituting them back into both original equations. This ensures that the solutions are correct and satisfy the entire system. Checking solutions helps to identify any errors made during the solving process and confirms the accuracy of the results.

By following these steps systematically, you can effectively solve a wide range of systems of equations using the substitution method. Now, let’s apply these steps to our specific system.

Solving the System: A Detailed Walkthrough

Let's apply the substitution method to the given system:

$\left\{\begin{aligned} x^2-4 x+y^2 & =-3 \\ x-y & =3 \end{aligned}\right.$

Step 1: Isolate a Variable

We'll start by isolating x in the second equation, x - y = 3. Adding y to both sides gives us:

x = y + 3

This expression is simple and easy to substitute, making it an ideal starting point for the substitution method. Isolating x in this manner provides a clear and direct way to express x in terms of y, which will be crucial for the next step.

Step 2: Substitute into the Other Equation

Now we substitute x = y + 3 into the first equation, x² - 4x + y² = -3. Replacing every instance of x with (y + 3) gives us:

(y + 3)² - 4(y + 3) + y² = -3

This substitution is the core of the method, transforming the system into a single equation with only one variable, y. The resulting equation may look complex, but it is now solvable using standard algebraic techniques. The next step will focus on simplifying and solving this equation.

Step 3: Solve the Resulting Equation

Expanding and simplifying the equation, we get:

(y² + 6y + 9) - 4y - 12 + y² = -3

Combining like terms, we have:

2y² + 2y - 3 = -3

Adding 3 to both sides, we get a quadratic equation:

2y² + 2y = 0

We can factor out a 2y:

2y(y + 1) = 0

Setting each factor equal to zero gives us two possible values for y:

2y = 0  =>  y = 0
y + 1 = 0  =>  y = -1

So, we have two solutions for y: y = 0 and y = -1. These values will be used in the next step to find the corresponding values of x. Finding multiple solutions is common in systems involving quadratic equations, and it highlights the importance of considering all possibilities.

Step 4: Substitute Back to Find the Other Variable

Now we substitute the values of y back into the equation x = y + 3 to find the corresponding values of x:

  • If y = 0, then:

    x = 0 + 3 = 3

  • If y = -1, then:

    x = -1 + 3 = 2

This gives us two potential solutions: (3, 0) and (2, -1). It's crucial to verify these solutions to ensure they satisfy both original equations. Substituting back allows us to complete the solution process and obtain the pairs of values that solve the system.

Step 5: Check Your Solutions

We must check both solutions in the original system of equations to ensure they are correct.

Checking (3, 0):

  • Equation 1: x² - 4x + y² = -3

    (3)² - 4(3) + (0)² = 9 - 12 + 0 = -3

    This checks out.

  • Equation 2: x - y = 3

    3 - 0 = 3

    This also checks out.

Checking (2, -1):

  • Equation 1: x² - 4x + y² = -3

    (2)² - 4(2) + (-1)² = 4 - 8 + 1 = -3

    This checks out.

  • Equation 2: x - y = 3

    2 - (-1) = 2 + 1 = 3

    This checks out as well.

Both solutions satisfy both equations, confirming their validity. Checking solutions is a critical step in the substitution method as it helps to catch any errors and ensures the accuracy of the final answer.

The Final Solution

The solutions to the system of equations are (3, 0) and (2, -1). These points represent the intersections of the curves defined by the two equations. By systematically applying the substitution method and rigorously checking our solutions, we have successfully solved the given system of equations.

Key Takeaways

  • The substitution method is a powerful technique for solving systems of equations, especially when one equation can be easily solved for one variable.
  • Isolating a variable, substituting the expression, solving the resulting equation, and substituting back are the core steps of this method.
  • Always check your solutions to ensure accuracy.
  • Systems involving quadratic equations may have multiple solutions, each of which must be verified.

By mastering the substitution method and understanding its applications, you can confidently tackle a wide range of algebraic problems and gain a deeper appreciation for the beauty and utility of mathematics.

Practice Problems

To solidify your understanding of the substitution method, try solving the following systems of equations:

  1. {y=x2−2x+1y=x−1\left\{\begin{aligned} y &= x^2 - 2x + 1 \\ y &= x - 1 \end{aligned}\right.
  2. {x2+y2=25y=x+1\left\{\begin{aligned} x^2 + y^2 &= 25 \\ y &= x + 1 \end{aligned}\right.
  3. {y=x2−4y=2x−1\left\{\begin{aligned} y &= x^2 - 4 \\ y &= 2x - 1 \end{aligned}\right.

Work through each problem step-by-step, and remember to check your solutions. Practice is key to mastering any mathematical technique, and these exercises will help you build confidence in using the substitution method.