91Ó°ÊÓ

When solving linear equations, one of the key steps is isolating the variable. In our exercise, the variable is \(x\).
To isolate \(x\), we need to move all other elements to the opposite side of the equation. This often involves reversing operations - for instance, if \(x\) is multiplied by \(-3\), we divide both sides of the equation by \(-3\).

Consider our equation: \(-3x = \frac{1}{4}\). To make \(x\) stand alone, divide both sides by \(-3\). This operation looks like this:
\[x = \frac{\frac{1}{4}}{-3}\]].
Dividing by \(-3\) effectively isolates \(x\), setting the stage for the next steps. It's crucial to perform the same operation on both sides of the equation to maintain balance and accuracy.

Simplifying fractions is an essential skill in algebra. When we simplify, we break down the fraction to its simplest form, making it easier to understand and work with.

From isolating the variable, we had: \[x = \frac{\frac{1}{4}}{-3}\].
To simplify this, treat the complex fraction as multiplication:\br>\[x = \frac{1}{4} \times \frac{1}{-3}\].

Now, multiply the numerators and denominators:
\[ x = \frac{1 \times 1}{4 \times (-3)} = \frac{1}{-12}\]
Here, the negative sign can be pulled out to give:
\[ x = -\frac{1}{12}\].
This simplified form makes it clear and easy to interpret the value of \(x\). Mastering fraction simplification will help you solve many mathematical problems efficiently.

Understanding negative multiplication is crucial for solving equations correctly. When you multiply or divide by negative numbers, remember these rules:

• Positive × Negative = Negative
• Negative × Negative = Positive

In our problem, we divided by a negative number which changed the sign. Initially, we had:
\(-3x = \frac{1}{4}\)
Upon isolating \(x\) by dividing both sides by \(-3\), it introduced a negative sign to the fraction:\br>\[x = \frac{\frac{1}{4}}{-3} = \frac{1}{-12}\]
Thus, negative multiplication confirmed that \[ x = -\frac{1}{12} \]
Understanding these multiplication rules ensures you interpret and simplify problems correctly, preserving the equation's balance.