Solve Systems Of Equations: A Step-by-Step Guide

by Rajiv Sharma 49 views

Hey everyone! Let's dive into solving systems of equations. This can seem tricky at first, but with a bit of practice, you'll be solving them like a pro. We're going to tackle a specific system today and walk through the steps to find the solution. So, grab your pencils and let's get started!

The Challenge: A System of Three Equations

Our main keyword here is solving systems of equations, so let's jump right in. We've got a system of three equations with three variables (x, y, and z). These are the equations we need to solve:

x+2y=−2y+2z=−2x−y−z=4 \begin{aligned} x + 2y &= -2 \\ y + 2z &= -2 \\ x - y - z &= 4 \end{aligned}

Our goal is to find the values of x, y, and z that satisfy all three equations simultaneously. This means the values we find, when plugged into each equation, will make the equation true. There are several methods to solve systems of equations, but we'll focus on using substitution and elimination to tackle this one. These methods help us simplify the equations and isolate the variables, making it easier to find their values. Remember, the key to successfully solving systems of equations lies in a methodical approach and careful execution of each step. Keep an eye out for opportunities to simplify and eliminate variables, and always double-check your work to avoid errors. Systems of equations pop up all over the place, from balancing chemical equations to optimizing business costs, so mastering this skill will definitely pay off in the long run. So, let's break down this system step by step and uncover the values of x, y, and z that fit perfectly into these equations. We're on a mission to make systems of equations less intimidating and more approachable for everyone, one solved problem at a time.

Step 1: Isolating a Variable

To kick things off in solving systems of equations, we want to isolate one variable in one of the equations. This means getting a variable all by itself on one side of the equation. Looking at our system, the first equation, x + 2y = -2, seems like a good place to start. It's relatively simple and doesn't have any coefficients other than 1 in front of the x. Let's isolate x in this equation. To do that, we'll subtract 2y from both sides of the equation:

x+2y−2y=−2−2y x + 2y - 2y = -2 - 2y

This simplifies to:

x=−2−2y x = -2 - 2y

Great! Now we have x expressed in terms of y. This is a crucial step because we can now substitute this expression for x in another equation. Substitution is a powerful technique in solving systems of equations because it allows us to reduce the number of variables in an equation, making it easier to solve. By isolating x, we've essentially created a bridge between the first equation and the others. This bridge will help us to connect the equations and ultimately find the values that satisfy them all. Remember, the goal here is to simplify the system. By strategically isolating a variable, we're setting ourselves up for success in the next steps. It's like picking the right tool for the job – isolating x in the first equation gives us a clean and manageable expression that we can use to our advantage. So, with x nicely isolated, we're ready to move on to the next stage of our equation-solving adventure. We're one step closer to cracking this system of equations!

Step 2: Substitution Time!

Now comes the fun part: substitution! We've got x = -2 - 2y, and we're going to plug this expression into another equation to eliminate x. A key aspect of solving systems of equations is using substitution effectively. Let's choose the third equation, x - y - z = 4, for this substitution. We'll replace the x in this equation with our expression (-2 - 2y):

(−2−2y)−y−z=4 (-2 - 2y) - y - z = 4

See what we did there? We swapped out x for its equivalent expression. Now we have an equation with only y and z variables. This is a big win because we've reduced the number of variables in this equation, making it simpler to work with. Let's simplify this equation further by combining like terms:

−2−2y−y−z=4 -2 - 2y - y - z = 4

−2−3y−z=4 -2 - 3y - z = 4

Now, let's isolate the terms with variables on one side by adding 2 to both sides:

−3y−z=6 -3y - z = 6

We now have a new equation, -3y - z = 6, which involves only y and z. This equation, along with our second original equation, y + 2z = -2, forms a smaller system of two equations with two variables. This is a significant step forward. By using substitution, we've transformed our original three-variable system into a more manageable two-variable system. This is a common strategy in solving systems of equations: reduce the complexity by eliminating variables one at a time. Each substitution brings us closer to the solution. So, with our new equation in hand, we're ready to tackle this two-variable system and continue our quest to find the values of x, y, and z. Keep up the great work!

Step 3: Solving the Two-Variable System

Okay, guys, we've narrowed it down! We now have a system of two equations with two variables, y and z. This is much easier to handle. Our equations are:

y+2z=−2−3y−z=6 \begin{aligned} y + 2z &= -2 \\ -3y - z &= 6 \end{aligned}

To solve this, we can use either substitution or elimination again. Let's go with elimination this time. The main keyword here is solving systems of equations. We want to eliminate one of the variables by making the coefficients of either y or z opposites. Looking at the equations, it seems easier to eliminate z. To do this, we'll multiply the first equation by 1 and the second equation by 2:

1∗(y+2z)=1∗(−2)−−>y+2z=−2 1 * (y + 2z) = 1 * (-2) --> y + 2z = -2

2∗(−3y−z)=2∗6−−>−6y−2z=12 2 * (-3y - z) = 2 * 6 --> -6y - 2z = 12

Now our system looks like this:

y+2z=−2−6y−2z=12 \begin{aligned} y + 2z &= -2 \\ -6y - 2z &= 12 \end{aligned}

Notice that the coefficients of z are now opposites (+2 and -2). This is exactly what we wanted! Now we can add the two equations together. When we add the left sides, the 2z and -2z will cancel out:

(y+2z)+(−6y−2z)=−2+12 (y + 2z) + (-6y - 2z) = -2 + 12

Simplifying, we get:

−5y=10 -5y = 10

Now we can easily solve for y by dividing both sides by -5:

y=−2 y = -2

Boom! We've found the value of y. This is a major breakthrough in solving systems of equations. Once you find one variable, the rest often fall into place more easily. We're now one step closer to the complete solution. Remember, the key to elimination is to strategically manipulate the equations so that one variable cancels out when you add them. We did this by multiplying the equations by appropriate constants to make the coefficients of z opposites. With y in hand, we're ready to move on and find the value of z. Keep the momentum going!

Step 4: Finding z

Alright! We've nailed down that y = -2. Now, to find z, we'll simply substitute this value back into one of the equations that involve both y and z. Our focus here remains on solving systems of equations. Let's use the second original equation, y + 2z = -2, since it looks pretty straightforward:

(−2)+2z=−2 (-2) + 2z = -2

We've replaced y with its value, -2. Now we just need to solve for z. First, add 2 to both sides of the equation:

2z=0 2z = 0

Next, divide both sides by 2:

z=0 z = 0

Fantastic! We've found that z = 0. This is another crucial piece of the puzzle in solving our system of equations. With both y and z in our toolbox, we're in a great position to find the last variable, x. Remember, substitution is a powerful tool that allows us to leverage the values we've already found. By plugging the value of y into the equation y + 2z = -2, we were able to isolate z and determine its value. This step-by-step approach is key to successfully navigating systems of equations. It's like following a treasure map – each value we find leads us closer to the final destination. So, with y and z securely in hand, we're ready to embark on the final leg of our journey and uncover the value of x. Keep the problem-solving spirit alive!

Step 5: Solving for x

We're almost there, guys! We know that y = -2 and z = 0. To find x, we can substitute these values into any equation that contains x. We're still focused on solving systems of equations, so let's choose the first original equation, x + 2y = -2, as it seems the simplest:

x+2(−2)=−2 x + 2(-2) = -2

We've replaced y with its value, -2. Now, let's simplify and solve for x:

x−4=−2 x - 4 = -2

Add 4 to both sides:

x=2 x = 2

Awesome! We've found that x = 2. We now have the values for all three variables: x = 2, y = -2, and z = 0. This is the solution to our system of equations. Remember, the key to success in solving systems of equations is a systematic approach. We started by isolating a variable, then used substitution and elimination to reduce the complexity of the system. Now, with all the variables solved, we're ready to take the final step and verify our solution.

Step 6: The Grand Check

Okay, the moment of truth! We need to make sure our solution, x = 2, y = -2, and z = 0, actually works in all three original equations. This is a crucial step in solving systems of equations – you always want to double-check your work to avoid any mistakes. Let's plug these values into each equation and see if they hold true.

Equation 1:

x+2y=−2 x + 2y = -2

2+2(−2)=−2 2 + 2(-2) = -2

2−4=−2 2 - 4 = -2

−2=−2 -2 = -2

Check! The first equation is satisfied.

Equation 2:

y+2z=−2 y + 2z = -2

(−2)+2(0)=−2 (-2) + 2(0) = -2

−2+0=−2 -2 + 0 = -2

−2=−2 -2 = -2

Check! The second equation is also satisfied.

Equation 3:

x−y−z=4 x - y - z = 4

2−(−2)−0=4 2 - (-2) - 0 = 4

2+2−0=4 2 + 2 - 0 = 4

4=4 4 = 4

Check! The third equation holds true as well. We've done it! Our solution satisfies all three equations. This confirms that our values for x, y, and z are correct. Verifying your solution is an essential practice in solving systems of equations. It provides peace of mind and ensures that you haven't made any algebraic errors along the way. It's like the final seal of approval on your problem-solving journey. So, with a successful check under our belts, we can confidently declare that we've solved the system of equations!

The Final Answer

So, after all that hard work, we've arrived at the solution! The correct answer is:

C. (2, -2, 0)

Key Takeaways for Solving Systems of Equations

  • Isolate: Pick an equation and isolate a variable. This sets the stage for substitution.
  • Substitute: Substitute the expression for the isolated variable into another equation to reduce the number of variables.
  • Eliminate: Use elimination to get rid of another variable. This often involves multiplying equations by constants to create opposite coefficients.
  • Solve: Once you have a single variable equation, solve for that variable.
  • Back-Substitute: Plug the values you find back into previous equations to solve for the remaining variables.
  • Check: Always, always, always check your solution in the original equations.

Solving systems of equations might seem daunting at first, but with practice and a clear strategy, you'll become a pro in no time! Keep these steps in mind, and remember to take it one step at a time. Happy equation-solving!