Master Substitution: Solving Systems Of Equations

by Alex Johnson 50 views

Mastering Substitution: Solving Systems of Equations

Hey there, math enthusiasts! Today, we're diving deep into a super useful technique for solving systems of linear equations: substitution. It's like a puzzle where you swap pieces around until you find the perfect fit. We'll be tackling a specific example to really nail this down, so grab your notebooks and let's get started!

Understanding the Substitution Method

The substitution method is a powerful algebraic technique used to solve a system of two or more linear equations. The core idea is to isolate one variable in one of the equations and then substitute that expression into the other equation. This process reduces the system of two equations with two variables into a single equation with just one variable, making it much easier to solve. Once you find the value of that single variable, you can plug it back into either of the original equations to find the value of the other variable. It’s a systematic approach that works wonders when one of the variables in your system is already isolated or can be easily isolated. We'll be using this exact principle to solve the following system:

-5y - 10 = x

and

x + 9y = -2

Notice how the first equation, -5y - 10 = x, already has x isolated on one side. This is a golden ticket for the substitution method! It means we already have an expression for x that we can directly substitute into the second equation. This makes our job significantly easier. If neither equation had a variable isolated, our first step would be to manipulate one of the equations to achieve this. For instance, if we had 2x + y = 5, we could easily isolate y by subtracting 2x from both sides, giving us y = 5 - 2x. This isolated expression for y would then be ready for substitution. The beauty of the substitution method lies in its ability to simplify complex systems into manageable single-variable equations. It's a fundamental skill in algebra that opens the door to solving more intricate problems in mathematics and various real-world applications, from economic modeling to engineering design. So, let's keep this first equation in mind, as it's our key to unlocking the solution.

Step-by-Step Solution

Let's break down the process of solving our system using substitution. Remember our equations:

  1. -5y - 10 = x
  2. x + 9y = -2

Step 1: Identify an Isolated Variable

As we observed, the first equation -5y - 10 = x conveniently provides an expression for x. This is precisely what we need for substitution.

Step 2: Substitute the Expression

Now, we take the expression for x from the first equation (-5y - 10) and substitute it into the second equation wherever we see x. So, in the equation x + 9y = -2, we replace x with (-5y - 10):

(-5y - 10) + 9y = -2

Step 3: Solve for the Remaining Variable

We now have a single equation with only one variable, y. Let's simplify and solve for y:

  • Combine like terms: -5y + 9y becomes 4y.
  • The equation is now: 4y - 10 = -2.
  • To isolate the y term, add 10 to both sides: 4y - 10 + 10 = -2 + 10.
  • This simplifies to: 4y = 8.
  • Finally, divide both sides by 4 to find the value of y: 4y / 4 = 8 / 4.
  • So, y = 2.

We've successfully found the value of y! This is a huge step. Keep in mind that the goal is to find the values of both x and y that satisfy both original equations simultaneously. Finding y is only half the battle, but it's often the most involved part of the substitution process. The algebraic manipulations here are crucial – combining like terms, isolating variables, and performing the same operation on both sides of the equation are fundamental skills that apply to countless mathematical problems. Take a moment to appreciate how the substitution transformed a two-variable problem into a simple one-variable equation. This efficiency is what makes the substitution method so powerful and widely used.

Step 4: Substitute the Found Value Back

Now that we know y = 2, we can substitute this value back into either of the original equations to find x. The first equation, -5y - 10 = x, looks like the easiest one to use because x is already isolated:

  • Substitute y = 2 into -5y - 10 = x:
  • -5(2) - 10 = x
  • Calculate: -10 - 10 = x
  • This gives us: x = -20.

And there you have it! We've found the value of x. It's always a good idea to double-check your work, especially when you're first learning a new method. You can do this by plugging both x = -20 and y = 2 into the other original equation (the one you didn't use for the final substitution) to see if it holds true.

Step 5: Verify the Solution

Let's check our solution (x = -20, y = 2) using the second original equation: x + 9y = -2.

  • Substitute the values: (-20) + 9(2) = -2.
  • Simplify: -20 + 18 = -2.
  • Calculate: -2 = -2.

Since the equation holds true, our solution is correct! The pair of values (x, y) = (-20, 2) is the unique solution to this system of equations. This verification step is critical. It's your final checkpoint to ensure that you haven't made any algebraic errors along the way. By substituting the calculated values back into both original equations (or at least the one not used in the final step), you confirm that these values simultaneously satisfy all conditions of the system. This process not only validates your answer but also reinforces your understanding of what it means for a pair of values to be a solution to a system of equations – they must make all equations in the system true statements. It's this meticulous verification that builds confidence and accuracy in mathematical problem-solving.

Conclusion

So, there you have it! We've successfully navigated the substitution method to solve a system of linear equations. We started with two equations, skillfully isolated a variable, substituted it into the other equation, solved for one variable, and then back-substituted to find the other. The key takeaway is that the substitution method simplifies systems by reducing the number of variables, making them much more approachable. Practice is your best friend when it comes to mastering algebraic techniques like substitution. The more systems you solve, the more comfortable you'll become with the steps and the quicker you'll be able to spot the most efficient way to solve them. Remember to always verify your solution by plugging your answers back into the original equations. This ensures accuracy and builds confidence in your mathematical abilities.

For further exploration into solving systems of equations and other algebraic concepts, you can check out resources like Khan Academy, a fantastic platform offering free lessons and practice problems.