Solving Systems Of Equations A Step-by-Step Guide

by Scholario Team 50 views

Hey guys! Today, we're diving into the fascinating world of solving systems of equations. This is a crucial skill in mathematics, and it pops up everywhere, from algebra to calculus and even in real-world applications. So, buckle up and let's get started!

Why are Systems of Equations Important?

Before we jump into the nitty-gritty, let's understand why solving systems of equations is so vital. Think of it this way: many real-life problems involve multiple unknowns and relationships between them. Systems of equations provide us with the tools to tackle these problems systematically.

For example, imagine you're trying to figure out the cost of two different items given the total cost of a combination of them. Or perhaps you're trying to determine the speeds of two cars traveling towards each other. These scenarios often translate into systems of equations, making them incredibly practical.

Solving systems of equations is not just about finding numbers; it's about uncovering hidden relationships and making informed decisions. This is why it's a core concept in various fields, including engineering, economics, and computer science.

Methods for Solving Systems of Equations

There are several methods we can use to solve systems of equations, each with its own strengths and weaknesses. We'll be focusing on two popular methods today: substitution and elimination. Let's briefly introduce them before we dive into specific examples.

1. Substitution Method

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This reduces the system to a single equation with one variable, which we can easily solve. Once we find the value of one variable, we can substitute it back into either of the original equations to find the value of the other variable. This method is particularly useful when one of the equations is already solved for one variable or can be easily rearranged.

2. Elimination Method

The elimination method (also known as the addition method) involves manipulating the equations so that the coefficients of one of the variables are opposites. Then, we add the equations together, which eliminates one variable and leaves us with a single equation in one variable. Again, we solve for the remaining variable and substitute back to find the other. This method shines when the coefficients of one of the variables are easily made opposites by multiplying one or both equations by a constant.

Let's Solve Some Systems of Equations!

Now that we've covered the basics, let's put our knowledge into action. We'll walk through several examples, showing you the step-by-step process for solving systems of equations using both substitution and elimination. Get ready to sharpen your pencils (or fire up your favorite equation solver)!

Example A: {x + 3y = 7 {5x - 7y = 31

Let's tackle this system using the elimination method. Our goal is to eliminate either x or y. Notice that if we multiply the first equation by -5, the coefficient of x will be -5, which is the opposite of the coefficient of x in the second equation.

  1. Multiply the first equation by -5: -5(x + 3y) = -5(7) => -5x - 15y = -35
  2. Now we have the following system:
    • {-5x - 15y = -35
    • {5x - 7y = 31
  3. Add the two equations together. Notice how the x terms cancel out:
    • (-5x - 15y) + (5x - 7y) = -35 + 31
    • -22y = -4
  4. Solve for y: y = -4 / -22 = 2/11
  5. Substitute the value of y (2/11) back into either of the original equations to solve for x. Let's use the first equation:
    • x + 3(2/11) = 7
    • x + 6/11 = 7
    • x = 7 - 6/11
    • x = (77 - 6) / 11
    • x = 71/11

Therefore, the solution to this system is x = 71/11 and y = 2/11.

Example B: {2x - y = 12 {3x + 2y = 25

For this system, let's use the substitution method. We can easily solve the first equation for y:

  1. Solve the first equation for y:
    • 2x - y = 12
    • -y = 12 - 2x
    • y = 2x - 12
  2. Substitute this expression for y into the second equation:
    • 3x + 2(2x - 12) = 25
  3. Simplify and solve for x:
    • 3x + 4x - 24 = 25
    • 7x = 49
    • x = 7
  4. Substitute the value of x (7) back into the equation we derived for y:
    • y = 2(7) - 12
    • y = 14 - 12
    • y = 2

Thus, the solution to this system is x = 7 and y = 2.

Example C: {x - y = -5 {2x + 3y = 10

Let's switch back to the elimination method for this one. We can eliminate x by multiplying the first equation by -2.

  1. Multiply the first equation by -2: -2(x - y) = -2(-5) => -2x + 2y = 10
  2. Now we have the following system:
    • {-2x + 2y = 10
    • {2x + 3y = 10
  3. Add the two equations together. The x terms cancel out:
    • (-2x + 2y) + (2x + 3y) = 10 + 10
    • 5y = 20
  4. Solve for y: y = 20 / 5 = 4
  5. Substitute the value of y (4) back into either original equation. Let's use the first equation:
    • x - 4 = -5
    • x = -5 + 4
    • x = -1

So, the solution to this system is x = -1 and y = 4.

Example D: {3x - y = -11 {x + 2y = 8

Let's use substitution again. We can easily solve the second equation for x:

  1. Solve the second equation for x:
    • x + 2y = 8
    • x = 8 - 2y
  2. Substitute this expression for x into the first equation:
    • 3(8 - 2y) - y = -11
  3. Simplify and solve for y:
    • 24 - 6y - y = -11
    • -7y = -35
    • y = 5
  4. Substitute the value of y (5) back into the equation we derived for x:
    • x = 8 - 2(5)
    • x = 8 - 10
    • x = -2

Therefore, the solution to this system is x = -2 and y = 5.

Example E: {2x + 3y = 9 {4x - 5y = 7

This system looks like a good candidate for elimination. Let's eliminate x. We can multiply the first equation by -2 to make the coefficients of x opposites.

  1. Multiply the first equation by -2: -2(2x + 3y) = -2(9) => -4x - 6y = -18
  2. Now we have the following system:
    • {-4x - 6y = -18
    • {4x - 5y = 7
  3. Add the two equations together. The x terms cancel out:
    • (-4x - 6y) + (4x - 5y) = -18 + 7
    • -11y = -11
  4. Solve for y: y = -11 / -11 = 1
  5. Substitute the value of y (1) back into either original equation. Let's use the first equation:
    • 2x + 3(1) = 9
    • 2x + 3 = 9
    • 2x = 6
    • x = 3

Thus, the solution to this system is x = 3 and y = 1.

Key Takeaways for Solving Systems of Equations

  • Understand the Goal: Remember that the goal is to find the values of the variables that satisfy all equations in the system.
  • Choose the Right Method: Consider the structure of the equations. Substitution works well when one equation is easily solved for a variable. Elimination is effective when coefficients can be easily made opposites.
  • Be Organized: Keep your work neat and organized to avoid errors. Write down each step clearly.
  • Check Your Solutions: Always substitute your solutions back into the original equations to verify that they work.

Practice Makes Perfect!

Solving systems of equations is a skill that improves with practice. The more you work through examples, the more comfortable you'll become with the different methods and techniques. So, grab some practice problems and start solving! You've got this!

By mastering solving systems of equations, you'll unlock a powerful tool for tackling a wide range of mathematical and real-world challenges. Keep practicing, stay curious, and happy solving, guys!