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How can I solve the equation?

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How can I solve the equation?

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$$If\ y\ne0\ and\ \frac{x}{y}=3,\ then\ \frac{\left(x-y\right)}{x}=\ ?$$

A. -2
B. -1
C. -2/3
D. 2/3
E. 1

The OA is D .

How can I use the hypothesis to solve the equation? It is not the same fraction.

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Vincen wrote:
$$If\ y\ne0\ and\ \frac{x}{y}=3,\ then\ \frac{\left(x-y\right)}{x}=\ ?$$

A. -2
B. -1
C. -2/3
D. 2/3
E. 1

The OA is D .

How can I use the hypothesis to solve the equation? It is not the same fraction.
Pick easy numbers. If x = 3 and y = 1, then (x-y)/x = (3-1)/3 = 2/3. The answer is D

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You have a few options here:

#1: Substitution
If x/y = 3, then x = 3y
So, given
$$\ \frac{\left(x-y\right)}{x}=\ ?$$
Rewrite as
$$\frac{3y-y}{3y}$$
$$\frac{2y}{3y}$$
$$\frac{2}{3}$$

The answer is D.

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Option #2: Split the fraction
If we have a single term in our denominator & we're adding or subtracting in the numerator, we're allowed to split it into 2 fractions:
$$\frac{x-y}{x}=\frac{x}{x}-\frac{y}{x}$$
$$1-\frac{y}{x}$$

If $$\frac{x}{y}=3$$
then $$\frac{y}{x}=\frac{1}{3}$$

So, $$1-\frac{1}{3}=\frac{2}{3}$$

The answer is D.

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Quote:
$$If\ y\ne0\ and\ \frac{x}{y}=3,\ then\ \frac{\left(x-y\right)}{x}=\ ?$$

A. -2
B. -1
C. -2/3
D. 2/3
E. 1

The OA is D .

How can I use the hypothesis to solve the equation? It is not the same fraction.
Hi Vincen,
Let's take a look at your question.

We will use the substitution method to solve it.
The first equation is :
$$\frac{x}{y}=3$$
Which implies:
$$x=3y\ $$

Plugin x = 3y in second equation.
$$\frac{\left(x-y\right)}{x}=?$$
$$=\frac{\left(3y-y\right)}{3y}$$
$$=\frac{2y}{3y}=\frac{2}{3}$$

Therefore, Option D is correct.

Hope it helps.
I am available if you'd like any follow up.

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Since x/y = 3, we know x = 3y.

Replacing x with with 3y in our other fraction gives

(x - y)/x =>

(3y - y)/3y =>

2y/3y =>

2/3

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