\frac{1}{4}\left(36x-8\right)=\frac{1}{2}\left(4x+24\right)

Let's solve the equation step by step:

\[
\frac{1}{4}(36x - 8) = \frac{1}{2}(4x + 24)
\]

### Step 1: Simplify both sides
On the left-hand side, distribute the \(\frac{1}{4}\) to both terms inside the parentheses:

\[
\frac{1}{4}(36x - 8) = \frac{1}{4}(36x) - \frac{1}{4}(8) = 9x - 2
\]

On the right-hand side, distribute the \(\frac{1}{2}\) to both terms inside the parentheses:

\[
\frac{1}{2}(4x + 24) = \frac{1}{2}(4x) + \frac{1}{2}(24) = 2x + 12
\]

Now the equation looks like this:

\[
9x - 2 = 2x + 12
\]

### Step 2: Move the variables to one side
Subtract \(2x\) from both sides to get all the terms with \(x\) on one side:

\[
9x - 2x - 2 = 12
\]

This simplifies to:

\[
7x - 2 = 12
\]

### Step 3: Isolate \(x\)
Add 2 to both sides to eliminate the constant on the left-hand side:

\[
7x = 14
\]

### Step 4: Solve for \(x\)
Divide both sides by 7:

\[
x = 2
\]

So, the solution is \(x = 2\).

Would you like further details or have any questions?

Here are 5 related questions to expand this topic:

1. How can you check if \(x = 2\) is the correct solution by substituting back into the original equation?
2. What happens if the coefficients on both sides are different? How would the solving method change?
3. How do you solve a similar equation with fractions, like \(\frac{1}{3}(9x - 15) = \frac{1}{6}(12x + 18)\)?
4. Can you solve an equation involving more complex distribution, like \(\frac{1}{5}(25x - 10) + 3 = \frac{2}{3}(6x + 9)\)?
5. How do you handle equations with variables on both sides that also have higher powers of \(x\)?

**Tip:** Always simplify the equation step by step to avoid mistakes, especially when distributing constants like \(\frac{1}{4}\) or \(\frac{1}{2}\).

\frac{1}{4}(36x - 8)=\frac{1}{2}(4x + 24)

Learn how to solve the linear equation with fractions \frac{1}{4}(36x - 8) = \frac{1}{2}(4x + 24) using a detailed step-by-step process. This problem covers algebraic concepts such as simplifying, distributing, and isolating the variable. Suitable for students in Grades 6-8.

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