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# Which of the following is equivalent to the pair of inequalities y > -

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Which of the following is equivalent to the pair of inequalities y > -  [#permalink]

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07 Feb 2017, 21:38
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Which of the following is equivalent to the pair of inequalities y > -3x and -z > 2x ?

(A) -2y < 6x < -3z
(B) 2y < -6x < -3z
(C) -3x < y < -z
(D) 3x < z < -y
(E) -3z < 12x < y

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Re: Which of the following is equivalent to the pair of inequalities y > -  [#permalink]

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07 Feb 2017, 21:50
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y > -3x --> Can be written as: -y < 3x --> Multiply both sides with 2 --> -2y < 6x --- Inequality A

-z > 2x ? --> Multiply both sides with 3 --> -3z > 6x --- Inequality B

Combining Inequalities A & B --> -2y < 6x < -3z

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Re: Which of the following is equivalent to the pair of inequalities y > -  [#permalink]

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07 Feb 2017, 21:58
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Which of the following is equivalent to the pair of inequalities y > -3x and -z > 2x ?

(A) -2y < 6x < -3z
(B) 2y < -6x < -3z
(C) -3x < y < -z
(D) 3x < z < -y
(E) -3z < 12x < y

y > -3x
or 3x > -y
multiplying 2 on both sides we get
6x > -2y

-z > 2x
or 2x < -z
multiplying 3 on both sides we get
6x < -3z

Combining both inequalities we get

-2y < 6x < -3z

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Re: Which of the following is equivalent to the pair of inequalities y > -  [#permalink]

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08 Feb 2017, 02:55
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Which of the following is equivalent to the pair of inequalities y > -3x and -z > 2x ?

(A) -2y < 6x < -3z
(B) 2y < -6x < -3z
(C) -3x < y < -z
(D) 3x < z < -y
(E) -3z < 12x < y

The first question might be what to do with these two inequalities - the solution is to look at the options.

Each variable is independent in the inequality. So we shouldn't combine the inequalities. We need the relation of x, y and z so we need to keep the inequalities separate. To compare them then we need one common term. That has to be the x term so make it equal in both inequalities (like you do in ratios)

You get 2y > -6x and -3z > 6x

Now note that z is negative in 4 of the 5 options so keep z negative. Multiply the first inequality by -1 to get

-2y < 6x and
-3z > 6x

We get -2y < 6x < -3z

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Re: Which of the following is equivalent to the pair of inequalities y > -  [#permalink]

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08 Feb 2017, 07:08
Which of the following is equivalent to the pair of inequalities y > -3x and -z > 2x ?

(A) -2y < 6x < -3z
(B) 2y < -6x < -3z
(C) -3x < y < -z
(D) 3x < z < -y
(E) -3z < 12x < y

(I) y > -3x
(II) -z > 2x

(I) * 2 = 2y > -6x
(II)* -3 = -6x > 3z

Now combine the two : $$2y > -6x > 3z$$

And if we multiply the inequality by -1 we have : $$-2y < 6x < -3z$$

Thus, answer will be (A) $$-2y < 6x < -3z$$
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Re: Which of the following is equivalent to the pair of inequalities y > -  [#permalink]

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08 Feb 2017, 07:16
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Which of the following is equivalent to the pair of inequalities y > -3x and -z > 2x ?

(A) -2y < 6x < -3z
(B) 2y < -6x < -3z
(C) -3x < y < -z
(D) 3x < z < -y
(E) -3z < 12x < y

We can also answer this question by examining only one part

Take -z > 2x, which we can rewrite as 2x < -z

Now examine the answer choices and focus solely on the relationships between x and z
(A) -2y < 6x < -3z
(B) 2y < -6x < -3z
(C) -3x < y < -z
(D) 3x < z < -y
(E) -3z < 12x < y

Can 2x < -z be rewritten to match any of these answer choices?
You bet.

Take 2x < -z and multiply both sides by 3 to get: 6x < -3z

If we look further, we see that 2x < -z CANNOT be rewritten to match any of the other 4 answer choices.
So, the correct answer must be A

Cheers,
Brent
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Re: Which of the following is equivalent to the pair of inequalities y > -  [#permalink]

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08 Feb 2017, 08:08
Which of the following is equivalent to the pair of inequalities y > -3x and -z > 2x ?

(A) -2y < 6x < -3z
(B) 2y < -6x < -3z
(C) -3x < y < -z
(D) 3x < z < -y
(E) -3z < 12x < y

We Cannot combine y > -3x and -z > 2x Since we need one common term

so multiple (y > -3x) * 2 = 2y > -6x & (-z > 2x)* -3 = -6x > 3z

As we have a common term -6x we can compare now. We Get -2y < 6x < -3z (A)
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Re: Which of the following is equivalent to the pair of inequalities y > -  [#permalink]

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02 Dec 2019, 02:00
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Re: Which of the following is equivalent to the pair of inequalities y > -   [#permalink] 02 Dec 2019, 02:00
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