Substitution Method: Solving Systems Of Equations

When you're faced with a system of linear equations, you've got a few tools in your mathematical toolbox to find the solution. One of the most straightforward and widely used methods is the substitution method. This technique is particularly handy when one of the equations can be easily rearranged to express one variable in terms of the other. It's like playing detective with your equations, isolating one clue (variable) to substitute it into another clue (equation) and uncover the hidden truth. We'll be diving deep into how this method works, breaking down the steps with a clear example:

$$\left\{\begin{array}{r} 9 x-5 y=-30 \\ x+3 y=18 \end{array}\right.$$

Get ready to unravel the mystery of these equations using the power of substitution!

Understanding the Substitution Method

The substitution method is a powerful algebraic technique used to solve systems of equations. At its core, it involves solving one of the equations for one variable, and then substituting that expression into the other equation. This reduces a system of two equations with two variables into a single equation with just one variable, which can then be solved. Once you've found the value of that variable, you can substitute it back into one of the original equations (or the rearranged one) to find the value of the second variable. It's a systematic approach that guarantees you'll find the unique solution where both equations are simultaneously true. Think of it like this: you have two pieces of information, and instead of trying to use them separately, you combine them to get a clearer picture. This method is especially elegant when one of the variables in one of the equations has a coefficient of 1 or -1, making it super easy to isolate.

Step-by-Step Guide to Substitution

Let's walk through the process of solving the system of equations using the substitution method. Our system is:

$$\left\{\begin{array}{r} 9 x-5 y=-30 \\ x+3 y=18 \end{array}\right.$$

Step 1: Isolate One Variable

The first step is to choose one of the equations and solve it for one of the variables. Look for the equation where it's easiest to get a variable by itself. In our system, the second equation, $x + 3y = 18$, looks like a great candidate because the coefficient of $x$ is 1. We can easily isolate $x$ by subtracting $3y$ from both sides:

$x = 18 - 3y$

Step 2: Substitute the Expression

Now that we have an expression for $x$ (which is $18 - 3y$), we're going to substitute this expression into the other equation. Remember, we used the second equation to isolate $x$, so we'll substitute into the first equation: $9x - 5y = -30$.

Replace every instance of $x$ in the first equation with $(18 - 3y)$:

$9(18 - 3y) - 5y = -30$

Step 3: Solve for the Remaining Variable

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

First, distribute the 9:

$162 - 27y - 5y = -30$

Combine the $y$ terms:

$162 - 32y = -30$

Now, isolate the term with $y$. Subtract 162 from both sides:

$-32y = -30 - 162$

$-32y = -192$

Finally, divide by -32 to find the value of $y$:

$y = \frac{-192}{-32}$

$y = 6$

Step 4: Substitute Back to Find the Other Variable

We've found the value of $y$! Now, we need to find the value of $x$. We can do this by substituting $y = 6$ back into either of the original equations, or, more conveniently, into the expression we found in Step 1 for $x$: $x = 18 - 3y$.

Substitute $y = 6$ into this expression:

$x = 18 - 3(6)$

$x = 18 - 18$

$x = 0$

Step 5: Check Your Solution

It's always a good idea to check your solution to make sure it satisfies both original equations. Our proposed solution is $(x, y) = (0, 6)$.

Let's check the first equation: $9x - 5y = -30$

$9(0) - 5(6) = -30$

$0 - 30 = -30$

$-30 = -30$ (This is true!)

Now let's check the second equation: $x + 3y = 18$

$(0) + 3(6) = 18$

$0 + 18 = 18$

$18 = 18$ (This is also true!)

Since our solution $(0, 6)$ satisfies both equations, it is the correct solution to the system.

When is the Substitution Method Most Useful?

The substitution method shines brightest when you encounter systems of equations where one variable in one of the equations has a coefficient of 1 or -1. This makes it incredibly simple to isolate that variable in a single step. For instance, if you see an equation like $y = 3x + 2$ or $x - 2y = 5$, the substitution method becomes very efficient. You can quickly express one variable in terms of the other and plug it into the second equation, streamlining the process. While substitution works for any system of linear equations, it's particularly advantageous when this condition is met, as it minimizes the algebraic manipulation needed in the initial steps. Even if no variable has a coefficient of 1, you can still use substitution, but it might involve working with fractions, which can sometimes be a bit more cumbersome than other methods like elimination. Therefore, always scan your system first to see if substitution offers a particularly clean path to the solution.

Alternatives to Substitution: The Elimination Method

While the substitution method is a fantastic tool, it's not the only way to solve systems of equations. The elimination method (also known as the addition method) offers an alternative approach that can be more efficient in certain situations. In the elimination method, the goal is to manipulate one or both equations by multiplying them by constants so that the coefficients of one of the variables are opposites. Then, by adding the two equations together, that variable is eliminated, leaving you with an equation in a single variable. For example, if you had the system:

$$\left\{\begin{array}{r} 2x + 3y = 7 \\ 4x - 3y = 5 \end{array}\right.$$

You can see that the $y$ coefficients are already opposites (+3y and -3y). Adding the equations directly would eliminate $y$: $(2x+4x) + (3y-3y) = 7+5$, leading to $6x = 12$, so $x=2$. This method is especially useful when no variable has a coefficient of 1, and it would be more work to isolate a variable for substitution.

Conclusion: Mastering Substitution for System Solutions

The substitution method is a fundamental and highly effective technique for solving systems of linear equations. By systematically isolating a variable in one equation and substituting its expression into the other, you can reduce the complexity of the system and efficiently find the values of each variable. Remember to always check your solution in both original equations to ensure accuracy. While other methods like elimination exist and have their own strengths, mastering substitution provides you with a reliable and versatile tool for tackling algebraic challenges. For further exploration into solving systems of equations and other algebraic concepts, you might find resources on Khan Academy or Math is Fun incredibly helpful.