Elimination Method: Solving Linear Equations

Let's dive into the world of solving linear equations using the elimination method! This technique is super handy when you have a system of two equations with two variables, and you want to find the values of those variables that satisfy both equations simultaneously. The core idea behind the elimination method is to manipulate one or both equations in such a way that when you add or subtract them, one of the variables cancels out (is eliminated), leaving you with a single equation with only one variable. This makes it much easier to solve! We'll walk through a specific example to make this crystal clear. Imagine you're presented with the following system of equations:

$$\begin{array}{l} 0.5 x-0.3 y=-2.4 \\ 0.3 x-0.5 y=-2.4 \end{array}$$

Our goal is to find the values of 'x' and 'y' that make both of these statements true. The elimination method is particularly elegant because it directly targets one of the variables. First, we need to decide which variable we want to eliminate. It doesn't really matter which one you choose; the process will lead you to the same correct answer. Let's say we decide to eliminate 'x' first. To do this, we need to make the coefficients of 'x' in both equations opposites. In our first equation, the coefficient of 'x' is 0.5, and in the second equation, it's 0.3. To make them opposites, we can multiply the first equation by 3 and the second equation by -5. Why these numbers? Because $0.5 \times 3 = 1.5$ and $0.3 \times -5 = -1.5$. See? They are now opposites!

So, let's perform these multiplications:

Equation 1 multiplied by 3:

$$3 \times (0.5 x - 0.3 y) = 3 \times (-2.4)$$

$$1.5x - 0.9y = -7.2$$

Equation 2 multiplied by -5:

$$-5 \times (0.3 x - 0.5 y) = -5 \times (-2.4)$$

$$-1.5x + 2.5y = 12$$

Now, look at our new system of equations:

$$\begin{array}{l} 1.5x - 0.9y = -7.2 \\ -1.5x + 2.5y = 12 \end{array}$$

Notice how the 'x' terms have coefficients that are opposites (1.5 and -1.5). This is exactly what we wanted! Now, we can add these two modified equations together. When we add them, the 'x' terms will cancel each other out:

$$(1.5x - 0.9y) + (-1.5x + 2.5y) = -7.2 + 12$$

$$(1.5x - 1.5x) + (-0.9y + 2.5y) = 4.8$$

$$0x + 1.6y = 4.8$$

$$1.6y = 4.8$$

And there you have it! We've successfully eliminated 'x' and are left with a simple equation with only 'y'. Now, we can easily solve for 'y' by dividing both sides by 1.6:

$$y = \frac{4.8}{1.6}$$

$$y = 3$$

So, we've found that $y = 3$. The next step is to find the value of 'x'. We can do this by substituting the value of 'y' (which is 3) back into either of the original equations. Let's use the first original equation: $0.5x - 0.3y = -2.4$.

Substitute $y = 3$:

$$0.5x - 0.3(3) = -2.4$$

$$0.5x - 0.9 = -2.4$$

Now, we need to isolate 'x'. First, add 0.9 to both sides of the equation:

$$0.5x = -2.4 + 0.9$$

$$0.5x = -1.5$$

Finally, divide both sides by 0.5 to find 'x':

$$x = \frac{-1.5}{0.5}$$

$$x = -3$$

And there we have it! We've found that $x = -3$. So, the solution to our system of equations is $x = -3$ and $y = 3$. It's always a good idea to check our answer by substituting these values back into both original equations to ensure they hold true.

Check with the first equation: $0.5x - 0.3y = -2.4$

$$0.5(-3) - 0.3(3) = -1.5 - 0.9 = -2.4$$

This is correct!

Check with the second equation: $0.3x - 0.5y = -2.4$

$$0.3(-3) - 0.5(3) = -0.9 - 1.5 = -2.4$$

This is also correct!

Since our values for 'x' and 'y' satisfy both equations, we can be confident that our solution, $x = -3$ and $y = 3$, is accurate. The elimination method provided a straightforward path to this solution by systematically removing one variable at a time. It’s a powerful tool in your algebra toolkit!

For more in-depth explanations and practice with solving systems of equations, you can explore resources like Khan Academy or Math is Fun.