Question

1. \\(\frac{3}{5}=\frac{\square}{\square}+\frac{\square}{\square}\\) \\(\frac{3}{5}=\frac{\square}{\square}+\frac{\square}{\square}+\frac{\square}{\square}\\)\
2. \\(1\frac{3}{4}=1+\frac{\square}{\square}\\) \\(1\frac{3}{4}=\frac{\square}{\square}+\frac{\square}{\square}\\)

Explanation:

Problem 3 (First Equation: $\boldsymbol{\frac{3}{5}=\frac{\square}{\square}+\frac{\square}{\square}}$)

Step1: Choose two fractions

We can split $\frac{3}{5}$ into two equal fractions. So, divide the numerator by 2: $\frac{3\div2}{5}=\frac{1.5}{5}$, but we want integer numerators. Alternatively, use $\frac{1}{5}+\frac{2}{5}$.
$\frac{1}{5}+\frac{2}{5}=\frac{1 + 2}{5}=\frac{3}{5}$

Step2: Verify

Adding $\frac{1}{5}$ and $\frac{2}{5}$ gives $\frac{3}{5}$, which matches the left - hand side.

Problem 3 (Second Equation: $\boldsymbol{\frac{3}{5}=\frac{\square}{\square}+\frac{\square}{\square}+\frac{\square}{\square}}$)

Step1: Split into three fractions

We can use three equal fractions. Divide the numerator 3 by 3: $\frac{3\div3}{5}=\frac{1}{5}$. So, $\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{1 + 1+1}{5}=\frac{3}{5}$

Step2: Verify

Adding $\frac{1}{5}$ three times: $\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{3}{5}$, which is correct.

Problem 4 (First Equation: $\boldsymbol{1\frac{3}{4}=1+\frac{\square}{\square}}$)

Step1: Recall mixed number definition

A mixed number $a\frac{b}{c}$ is equal to $a+\frac{b}{c}$. For $1\frac{3}{4}$, by the definition of mixed numbers, $1\frac{3}{4}=1+\frac{3}{4}$

Step2: Verify

Adding 1 and $\frac{3}{4}$ gives $1\frac{3}{4}$, which is correct.

Problem 4 (Second Equation: $\boldsymbol{1\frac{3}{4}=\frac{\square}{\square}+\frac{\square}{\square}}$)

Answer:

s:

1. $\frac{3}{5}=\frac{1}{5}+\frac{2}{5}$ (other valid answers like $\frac{3}{10}+\frac{3}{10}$ also exist); $\frac{3}{5}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}$ (other valid answers like $\frac{1}{5}+\frac{2}{10}+\frac{3}{10}$ also exist)
2. $1\frac{3}{4}=1+\frac{3}{4}$; $1\frac{3}{4}=\frac{4}{4}+\frac{3}{4}$ (other valid answers like $\frac{1}{4}+\frac{6}{4}$ also exist)