91Ó°ÊÓ

When solving linear equations, isolating the variable terms is the first crucial step. This means getting the variable (like \(x\) in this exercise) by itself on one side of the equation. Here's why: to solve an equation, we need to simplify it as much as possible.
Let's consider the example \(4x + 9 = 0\).
We want to isolate \(4x\), the term with the variable. To do this, you perform operations that will cancel out the other numbers or terms around it.
Here's how we do it:

• Subtract 9 from both sides: \(4x + 9 - 9 = 0 - 9\)
• This simplifies to: \(4x = -9\)

Notice how we moved the 9 to the other side. This is isolating the variable term. Now we can move on to the next step.
This is essential because it simplifies our task of solving for the variable.

After isolating the variable term, our task is to solve for the variable. This involves simplifying the equation further until we find what the variable equals.
Continuing with our example: \(4x = -9\).
To solve for \(x\), we need to get \(x\) by itself.
We do this by performing operations that reverse what's currently happening to \(x\).
In our example, \(x\) is being multiplied by 4. To reverse this, we divide both sides of the equation by 4:
\[ x = \frac{-9}{4} \]

• This gives us: \(x = -\frac{9}{4}\)

Now we've solved for the variable \(x\). Always check your work to ensure everything was simplified correctly. Solving for variables typically involves basic arithmetic operations, making it straightforward once you isolate the variable term first.

Factoring expressions is breaking down an equation into simpler components that can be multiplied together to get the original expression. It's an effective tool for simplifying and solving more complex equations.
Our exercise, \(4x + 9 = 0\), doesn't need factoring because it's already in a simple form.
However, understanding how to factor is important for more complex problems.
For example, consider the expression \(x^2 + 5x + 6\).
To factor it, we look for two numbers that multiply to 6 (the constant term) and add to 5 (the coefficient of the middle term).
These numbers are 2 and 3:

• So, \(x^2 + 5x + 6\) factors to \((x + 2)(x + 3)\)

Using this approach will help solve more advanced equations, making the process cleaner and more manageable. Always practice factoring simpler expressions to build a strong foundation for handling complex ones.