Solving Systems Of Equations Step By Step Guide

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Hey guys! Let's dive into the fascinating world of systems of equations. If you've ever wondered how to find the common solution that satisfies multiple equations at once, you're in the right place. We're going to break down a specific example, but the techniques we'll cover are applicable to a wide range of problems. So, grab your pencils, and let's get started!

The Problem: Our System of Equations

Our mission, should we choose to accept it, is to solve the following system of equations:

y = -3x - 2
5x + 2y = 15

What this basically means is that we need to find values for x and y that make both of these equations true simultaneously. Think of it like finding the exact point where two lines intersect on a graph. That point's x and y coordinates will be our solution.

Method 1: Substitution - A Clever Swap

One of the most powerful techniques for solving systems of equations is called substitution. The idea is super simple: if we know what one variable equals in terms of the other, we can substitute that expression into the other equation. This eliminates one variable and leaves us with a single equation we can solve.

Looking at our system of equations, we notice something awesome: the first equation, y = -3x - 2, already tells us what y is equal to! This is perfect for substitution. We can take the expression -3x - 2 and substitute it in place of y in the second equation.

Here’s how it looks:

  1. Start with the second equation:
    5x + 2y = 15
  2. Substitute y with -3x - 2:
    5x + 2(-3x - 2) = 15
    Notice how we replaced y with its equivalent expression. Now we have one equation with just one variable, x.
  3. Distribute and simplify:
    5x - 6x - 4 = 15
    We distributed the 2 across the parentheses.
  4. Combine like terms:
    -x - 4 = 15
    We combined the x terms.
  5. Isolate x:
    -x = 19
    We added 4 to both sides.
  6. Solve for x:
    x = -19
    We multiplied both sides by -1.

Woohoo! We've found the value of x: x = -19. But we're not done yet. We still need to find y.

Now that we know x, we can plug it back into either of the original equations to solve for y. The first equation, y = -3x - 2, looks easier, so let's use that one. This is the magic of substitution – you can always back-substitute to find the remaining variables.

  1. Start with the first equation:
    y = -3x - 2
  2. Substitute x with -19:
    y = -3(-19) - 2
  3. Simplify:
    y = 57 - 2
  4. Solve for y:
    y = 55

Awesome! We found y = 55. So, the solution to our system of equations is x = -19 and y = 55. This means the point (-19, 55) is the intersection point of the two lines represented by our equations.

To be absolutely sure we got the right answer, it's always a good idea to check our solution by plugging the x and y values back into both original equations. If both equations hold true, we know we've nailed it. Let's do that now!

  1. Check in the first equation:
    y = -3x - 2
55 = -3(-19) - 2
55 = 57 - 2
55 = 55  (This is true!)
  2. Check in the second equation:
    5x + 2y = 15
5(-19) + 2(55) = 15
-95 + 110 = 15
15 = 15  (This is also true!)

Both equations hold true, so we can confidently say that our solution, x = -19 and y = 55, is correct.

Method 2: Elimination - The Art of Vanishing Variables

Now, let's explore another powerful technique for tackling systems of equations: elimination. This method is also sometimes called the addition method because, as you'll see, it involves adding the equations together in a clever way to eliminate one of the variables. It’s a fantastic tool to have in your math arsenal.

The core idea behind elimination is to manipulate the equations so that the coefficients (the numbers in front of the variables) of one of the variables are opposites. When we add the equations together, that variable will vanish, leaving us with a single equation in one variable.

Let’s revisit our system of equations:

y = -3x - 2
5x + 2y = 15

Before we can start eliminating, we need to do a little rearranging. It’s helpful to have the x and y terms lined up on the same side of the equation. So, let's rewrite the first equation:

  1. Rewrite the first equation:
    y = -3x - 2
3x + y = -2
    We added 3x to both sides.

Now our system of equations looks like this:

3x + y = -2
5x + 2y = 15

Okay, now we need to figure out how to make the coefficients of either x or y opposites. Looking at the y terms, we see a y in the first equation and a 2y in the second. If we multiply the entire first equation by -2, the y term will become -2y, which is the opposite of 2y. This is our ticket to elimination!

  1. Multiply the first equation by -2:
    -2(3x + y) = -2(-2)
-6x - 2y = 4
    Remember to multiply every term in the equation by -2.

Now our system of equations looks like this:

-6x - 2y = 4
5x + 2y = 15

The moment we've been waiting for! We're ready to add the equations together. Notice what happens to the y terms:

  1. Add the equations:
    (-6x - 2y) + (5x + 2y) = 4 + 15
  2. Simplify:
    -x = 19
    The y terms canceled out! This is the magic of elimination.
  3. Solve for x:
    x = -19
    We multiplied both sides by -1.

Just like with substitution, we've found x = -19. Now we need to find y. We can plug this value of x back into either of the original equations to solve for y. Let’s use the first equation, y = -3x - 2, again.

  1. Start with the first equation:
    y = -3x - 2
  2. Substitute x with -19:
    y = -3(-19) - 2
  3. Simplify:
    y = 57 - 2
  4. Solve for y:
    y = 55

We got y = 55 again! This confirms our solution: x = -19 and y = 55. The elimination method led us to the same answer as substitution, which is exactly what we expect.

And, just like before, let's double-check our answer by plugging the x and y values back into both original equations. We already did this in the substitution section, and we know it works. So, we can be extra confident that our solution is correct.

Which Method is Best? Substitution vs. Elimination

So, we've explored two fantastic methods for solving systems of equations: substitution and elimination. You might be wondering, which one is the best? Well, the truth is, there’s no single “best” method. It often depends on the specific system of equations you're dealing with.

  • Substitution: This method shines when one of the equations is already solved for one variable (like our first equation, y = -3x - 2). It’s also a great choice if you can easily isolate one variable in one of the equations.
  • Elimination: This method is particularly useful when the coefficients of one of the variables are opposites or can easily be made opposites by multiplying one or both equations by a constant. It’s also a solid choice when the equations are in standard form (Ax + By = C).

The best approach is to become comfortable with both methods and then choose the one that seems most efficient for the given problem. Sometimes, you might even find that using a combination of the two methods is the most effective strategy. The more you practice, the better you'll become at recognizing which method is the best fit.

Wrapping Up: You're a System Solver!

Solving systems of equations is a fundamental skill in mathematics, and it has applications in many real-world scenarios. From calculating break-even points in business to modeling physical systems in science and engineering, the ability to find common solutions to multiple equations is incredibly valuable.

We've covered two powerful methods: substitution and elimination. Remember, the key is to practice, practice, practice! The more problems you solve, the more confident you'll become in your ability to tackle any system of equations that comes your way.

So, go forth and conquer those equations, guys! You've got this!