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If x + z > y + z, then which of the following must be true?
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28 Mar 2019, 05:59
28 Mar 2019, 05:59
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If \(x + z > y + z\), then which of the following must be true?
(I) \(x – z > y – z\)
(II) \(xz > yz\)
(III) \(\frac{x}{z} > \frac{y}{z}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only
(I) \(x – z > y – z\)
(II) \(xz > yz\)
(III) \(\frac{x}{z} > \frac{y}{z}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only
Re: If x + z > y + z, then which of the following must be true?
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30 Mar 2019, 12:09
30 Mar 2019, 12:09
1
Expert Reply
We can always add or subtract from both sides of an inequality. So if we just subtract z from both sides, we'll get x > y.
For statement (I), we can just add z to both sides to also get x > y. So that is definitely true.
For statements (II) and (III), though, there's a kick. We can only multiply or divide by a variable if we know its sign. This is because if we multiply or divide both sides of an inequality by a negative number, then the direction of the inequality flips. For example, 4 > 2, but if we multiply both sides by –2, then we get -8 < -4.
Since we don't know if z is positive or negative, we can't multiply or divide both sides by it. We don't know if we'd have to flip the sign or not. Hence, we don't know if (II) and (III) are true or not. If z is positive then they are true, but if z is negative, they are not.
For statement (I), we can just add z to both sides to also get x > y. So that is definitely true.
For statements (II) and (III), though, there's a kick. We can only multiply or divide by a variable if we know its sign. This is because if we multiply or divide both sides of an inequality by a negative number, then the direction of the inequality flips. For example, 4 > 2, but if we multiply both sides by –2, then we get -8 < -4.
Since we don't know if z is positive or negative, we can't multiply or divide both sides by it. We don't know if we'd have to flip the sign or not. Hence, we don't know if (II) and (III) are true or not. If z is positive then they are true, but if z is negative, they are not.
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If x + z > y + z, then which of the following must be true?
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31 Mar 2019, 07:20
31 Mar 2019, 07:20
2
Carcass wrote:
If \(x + z > y + z\), then which of the following must be true?
(I) \(x – z > y – z\)
(II) \(xz > yz\)
(III) \(\frac{x}{z} > \frac{y}{z}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only
(I) \(x – z > y – z\)
(II) \(xz > yz\)
(III) \(\frac{x}{z} > \frac{y}{z}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only
Let's check each statement
(I) x – z > y – z
It is given that: x + z > y + z
If we subtract 2z from both sides, we get: x - z > y - z
Perfect!
Statement I must be true.
Check the answer choices . . . ELIMINATE B, C and E since they suggest that statement 1 is NOT true.
(II) xz > yz
This statement need not be true.
It is given that: x + z > y + z
So, one possible case is that x = 5, y = 4 and z = -1, since 5 + (-1) > 4 + (-1)
When we plug these values into Statement II, we get: (5)(-1) > (4)(-1)
Simplify to get: -5 > -4, which is NOT TRUE
So, Statement II is not true
Check the remaining answer choices . . . ELIMINATE D
By the process of elimination, the correct answer is A
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