$$If\ y\ne0\ and\ \frac{x}{y}=3,\ then\ \frac{\left(x-y\right)}{x}=\ ?$$
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.
How can I solve the equation?
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- DavidG@VeritasPrep
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Pick easy numbers. If x = 3 and y = 1, then (x-y)/x = (3-1)/3 = 2/3. The answer is DVincen 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.
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- ceilidh.erickson
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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.
#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.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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- ceilidh.erickson
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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.
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.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
- EconomistGMATTutor
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Hi Vincen,$$If\ y\ne0\ and\ \frac{x}{y}=3,\ then\ \frac{\left(x-y\right)}{x}=\ ?$$
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.
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
Replacing x with with 3y in our other fraction gives
(x - y)/x =>
(3y - y)/3y =>
2y/3y =>
2/3