The World of Numbers | Exercise 3.3

Question 6

Show that:

$\left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3} = \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3}$

Solution

Understand the Question

This problem demonstrates the distributive property of multiplication over addition: $(a + b) \times c = a \times c + b \times c$ .
To show that both sides are equal, we evaluate them independently:
- Left Hand Side (LHS): Add the fractions inside the bracket first, then multiply by $\dfrac{8}{3}$ .
- Right Hand Side (RHS): Perform each multiplication separately, then add the resulting fractions.
If both calculations result in the same value, the equality is verified.

Step 1 · Calculate the Left Hand Side (LHS)

First, add the fractions inside the bracket using a common denominator of $4$ : Diagram 1

\begin{aligned} \dfrac{1}{2} + \dfrac{3}{4} &= \dfrac{1 \times 2}{2 \times 2} + \dfrac{3}{4} \\[0.6em] &= \dfrac{2}{4} + \dfrac{3}{4} \\[0.6em] &= \dfrac{2+3}{4} \\[0.6em] &= \dfrac{5}{4} \end{aligned}

Now, multiply by $\dfrac{8}{3}$ :

\begin{aligned} \text{LHS} &= \left(\dfrac{5}{4}\right) \times \dfrac{8}{3} \\[0.6em] &= \dfrac{5 \times 8}{4 \times 3} \\[0.6em] &= \dfrac{40}{12} \\[0.6em] &= \dfrac{10}{3} \quad \dots (i) \end{aligned}

Step 2 · Calculate the Right Hand Side (RHS)

Evaluate each multiplication term separately:

\begin{aligned} \dfrac{1}{2} \times \dfrac{8}{3} &= \dfrac{1 \times 8}{2 \times 3} \\[0.6em] &= \dfrac{8}{6} \\[0.6em] &= \dfrac{4}{3} \end{aligned}

\begin{aligned} \dfrac{3}{4} \times \dfrac{8}{3} &= \dfrac{3 \times 8}{4 \times 3} \\[0.6em] &= \dfrac{24}{12} \\[0.6em] &= 2 \end{aligned}

Now, add the two terms:

\begin{aligned} \text{RHS} &= \dfrac{4}{3} + 2 \\[0.6em] &= \dfrac{4}{3} + \dfrac{2 \times 3}{3} \\[0.6em] &= \dfrac{4}{3} + \dfrac{6}{3} \\[0.6em] &= \dfrac{4+6}{3} \\[0.6em] &= \dfrac{10}{3} \quad \dots (ii) \end{aligned}

From $(i)$ and $(ii)$ , $\text{LHS} = \text{RHS}$ .

Answer

Since $\text{LHS} = \text{RHS} = \dfrac{10}{3}$ , the statement is verified: $\left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3} = \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3}$

Common Mistakes

Adding Denominators Directly: Adding fractions incorrectly by adding tops and bottoms directly (e.g. $\dfrac{1}{2} + \dfrac{3}{4} \neq \dfrac{1+3}{2+4} = \dfrac{4}{6}$ ). Always find the common denominator first.
Order of Operations: In RHS, adding before multiplying violates the order of operations (BODMAS/PEMDAS). Each multiplication must be evaluated before performing the addition.
Incomplete Simplification: Leaving fractions unsimplified (such as $\dfrac{40}{12}$ or $\dfrac{8}{6}$ ), which makes comparing LHS and RHS more difficult.

Question 6

More questions in Exercise 3.3