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Isolating variables is a fundamental concept in solving equations. When we solve an equation, we often want to find the value of one specific variable. To do this, we need to 'isolate' that variable on one side of the equation.

This means that we want to get the variable by itself. For example, consider the equation: \[4x + 3y = 8\].

To isolate the variable 'x,' we first subtract '3y' from both sides. Why? Because we want to move the '3y' term away from the '4x.'

The subtraction leaves us with: \[4x = 8 - 3y\].

Now, 'x' is almost isolated. There is just one more step!

Algebraic manipulation involves using operations like addition, subtraction, multiplication, and division to rearrange equations. It is crucial for isolating variables and solving equations.

Let's continue from where we left off: \[4x = 8 - 3y\].

Here, we need to perform algebraic manipulation to solve for 'x.' We need to get rid of the '4' that is multiplied by 'x.' How do we do this? By dividing both sides by '4.' This will help us simplify the equation.

When we divide the left side by 4, we're left with: \[ \frac{4x}{4} = x \]

Dividing the right side by '4,' we get: \[ \frac{8 - 3y}{4}\].

So, our equation simplifies to: \[x = \frac{8 - 3y}{4}\].

Now, 'x' is by itself! Algebraic manipulation allowed us to isolate 'x' successfully.

Solving for 'x' means finding the value of 'x' that satisfies the equation. We did this by isolating 'x' using algebraic manipulation.

Let's summarize the steps we took:

• First, we started with the equation: \[4x + 3y = 8\].
• We wanted to solve for 'x,' so we needed to isolate it. We subtracted '3y' from both sides, which gave us: \[4x = 8 - 3y\].
• Then, we performed algebraic manipulation by dividing both sides by '4.' This simplified the equation to: \[x = \frac{8 - 3y}{4}\].

Notice that solving for 'x' involves understanding both isolating the variable and using correct algebraic manipulations. We moved terms around and simplified the equation until we found 'x.'

And that's it! By following these steps, you can solve for 'x' in any similar linear equation.