https://r2.erweima.ai/mathbot/image/1726505420907nykuwp22.jpg

Let's work through the first part of the image, focusing on **Exercise 8.1**.

### 1. Find the value of the unknown variables

#### a) \(\frac{-s}{24} = \frac{7}{6}\)

Multiply both sides by 24 to isolate \(s\):

\[
-s = \frac{7}{6} \times 24
\]
\[
-s = 28 \quad \Rightarrow \quad s = -28
\]

#### b) \(\frac{2.0x}{0.032} = -100\)

Multiply both sides by 0.032:

\[
2.0x = -100 \times 0.032
\]
\[
2.0x = -3.2
\]

Now divide both sides by 2:

\[
x = \frac{-3.2}{2} = -1.6
\]

#### c) \(\frac{6}{3x} + \frac{10}{3x} = 256\)

Since both terms have the same denominator, combine them:

\[
\frac{6 + 10}{3x} = 256 \quad \Rightarrow \quad \frac{16}{3x} = 256
\]

Multiply both sides by \(3x\) and divide by 256:

\[
16 = 256 \times 3x \quad \Rightarrow \quad 3x = \frac{16}{256} = \frac{1}{16}
\]
\[
x = \frac{1}{16 \times 3} = \frac{1}{48}
\]

#### d) \( \frac{1 - x}{3} - \frac{3}{2}\left(\frac{2 - x}{2}\right) + \frac{x + 3}{4} = 1 \)

This is a more complex equation involving fractions. We'll simplify this step-by-step.

Let's simplify inside the parentheses first:

\[
\frac{3}{2} \times \frac{2 - x}{2} = \frac{3(2 - x)}{4} = \frac{6 - 3x}{4}
\]

Now substitute back into the equation:

\[
\frac{1 - x}{3} - \frac{6 - 3x}{4} + \frac{x + 3}{4} = 1
\]

We can combine the second and third terms on the left-hand side since they have the same denominator:

\[
\frac{1 - x}{3} - \frac{6 - 3x - (x + 3)}{4} = 1 \quad \Rightarrow \quad \frac{1 - x}{3} - \frac{6 - 3x - x - 3}{4} = 1
\]
\[
\frac{1 - x}{3} - \frac{3 - 4x}{4} = 1
\]

Now, multiply through by 12 to eliminate the denominators:

\[
4(1 - x) - 3(3 - 4x) = 12
\]

Expanding both sides:

\[
4 - 4x - 9 + 12x = 12
\]
\[
-4x + 12x = 12 - 4 + 9
\]
\[
8x = 17 \quad \Rightarrow \quad x = \frac{17}{8}
\]

We have now solved the first four problems. Do you want to continue with the rest, or should I explain further details about these solutions?

### Follow-up Questions:
1. How would the solutions change if any negative sign was misplaced in part (d)?
2. Can you solve the remaining parts (e) to (i) on your own, or would you like assistance with those too?
3. In part (c), how would the equation change if the fractions had different denominators?
4. Why did multiplying by 12 help simplify the equation in part (d)?
5. What steps would you take if part (d) had decimals instead of fractions?

### Tip:
Always try to simplify complex fraction equations by clearing the denominators using multiplication before solving for the variable.

Find the value of the unknown variables and solve the following equations in Exercise 8.1.

Learn how to solve linear equations involving fractions from Exercise 8.1. This guide walks through the step-by-step solutions for variables in algebraic expressions, covering essential concepts such as combining fractions and isolating variables. Suitable for Grades 8-10.

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