Solve question 7 in the picture with stes

Sure, let's solve each part of question 7 step-by-step from the image.

### Question 7(a)

**If \( \frac{a}{b} = \frac{c}{d} \), prove that \( \frac{b}{a} = \frac{d}{c} \).**

Given:

\[
\frac{a}{b} = \frac{c}{d}
\]

This implies:

\[
a \cdot d = b \cdot c
\]

We need to prove:

\[
\frac{b}{a} = \frac{d}{c}
\]

Taking the reciprocals of both sides of the given equation:

\[
\frac{b}{a} = \frac{d}{c}
\]

Hence, it is proved.

### Question 7(b)

**If \( \frac{a}{b} = \frac{c}{d} \), prove that \( \frac{a + b}{b} = \frac{c + d}{d} \).**

Given:

\[
\frac{a}{b} = \frac{c}{d}
\]

This implies:

\[
a = \frac{b \cdot c}{d}
\]

We need to prove:

\[
\frac{a + b}{b} = \frac{c + d}{d}
\]

Starting with the left-hand side (LHS):

\[
\frac{a + b}{b}
\]

Substitute \( a \):

\[
\frac{\frac{b \cdot c}{d} + b}{b}
\]

Factor out \( b \):

\[
\frac{b \left( \frac{c}{d} + 1 \right)}{b}
\]

Simplify:

\[
\frac{c}{d} + 1 = \frac{c + d}{d}
\]

So, we have:

\[
\frac{a + b}{b} = \frac{c + d}{d}
\]

Hence, it is proved.

### Question 7(c)

**If \( \frac{a}{b} = \frac{c}{d} \), prove that \( \frac{a - b}{b} = \frac{c - d}{d} \).**

Given:

\[
\frac{a}{b} = \frac{c}{d}
\]

This implies:

\[
a = \frac{b \cdot c}{d}
\]

We need to prove:

\[
\frac{a - b}{b} = \frac{c - d}{d}
\]

Starting with the left-hand side (LHS):

\[
\frac{a - b}{b}
\]

Substitute \( a \):

\[
\frac{\frac{b \cdot c}{d} - b}{b}
\]

Factor out \( b \):

\[
\frac{b \left( \frac{c}{d} - 1 \right)}{b}
\]

Simplify:

\[
\frac{c}{d} - 1 = \frac{c - d}{d}
\]

So, we have:

\[
\frac{a - b}{b} = \frac{c - d}{d}
\]

Hence, it is proved.

### Question 7(d)

**If \( \frac{a}{b} = \frac{c}{d} \), prove that \( \frac{a + b}{a - b} = \frac{c + d}{c - d} \).**

Given:

\[
\frac{a}{b} = \frac{c}{d}
\]

This implies:

\[
a = \frac{b \cdot c}{d}
\]

We need to prove:

\[
\frac{a + b}{a - b} = \frac{c + d}{c - d}
\]

Starting with the left-hand side (LHS):

\[
\frac{a + b}{a - b}
\]

Substitute \( a \):

\[
\frac{\frac{b \cdot c}{d} + b}{\frac{b \cdot c}{d} - b}
\]

Factor out \( b \):

\[
\frac{b \left( \frac{c}{d} + 1 \right)}{b \left( \frac{c}{d} - 1 \right)}
\]

Simplify:

\[
\frac{\frac{c}{d} + 1}{\frac{c}{d} - 1}
\]

Rewrite:

\[
\frac{\frac{c + d}{d}}{\frac{c - d}{d}} = \frac{c + d}{c - d}
\]

So, we have:

\[
\frac{a + b}{a - b} = \frac{c + d}{c - d}
\]

Hence, it is proved.

If you have any questions or need further details, please let me know!

Here are five questions you might find interesting:
1. How do you convert between different units of measurement in scaling problems?
2. Can you explain the properties of proportions in more detail?
3. How do you solve more complex geometric scaling problems?
4. What are some applications of proportions in real life?
5. How can you simplify algebraic expressions involving proportions?

Tip: When working with proportions, always cross-multiply to check your work and ensure that the ratios are equivalent.

Explore how to solve Question 7 from the image, focusing on proportions in mathematics. Step-by-step solutions are provided for parts (a), (b), (c), and (d). Learn the principles of proportionality and their applications in mathematical proofs. Suitable for high school students.

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