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Question

Answer

$$1-\dfrac{x}{y}=\dfrac{-x+y}{y}$$

Explanation
$$ \begin{aligned}1-\frac{x}{y}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-x+y}{y}\end{aligned} $$
①

Subtract $ \dfrac{x}{y} $ from $ 1 $ to get $ \dfrac{ \color{purple}{ -x+y } }{ y }$.

Step 1: Write $ 1 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ y }$.

$$ \begin{aligned} 1- \frac{x}{y} & \xlongequal{\text{Step 1}} \frac{1}{\color{red}{1}} - \frac{x}{y} = \frac{ 1 \cdot \color{blue}{ y }}{ 1 \cdot \color{blue}{ y }} - \frac{ x }{ y } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ y } }{ y } - \frac{ \color{purple}{ x } }{ y }=\frac{ \color{purple}{ -x+y } }{ y } \end{aligned} $$
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