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4(x+2) < 2(x+5) + 14
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16 Apr 2021, 07:38
16 Apr 2021, 07:38
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Question Stats:
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1) \(4(x+2) \leq 2(x+5) + 14\)
2)\( \sqrt{16-8y+y^2} \leq 9\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
2)\( \sqrt{16-8y+y^2} \leq 9\)
Quantity A |
Quantity B |
x , where x is a solution of inequality 1 |
y, where y is a solution of inequality 2 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
4(x+2) < 2(x+5) + 14
[#permalink]
16 Apr 2021, 21:55
16 Apr 2021, 21:55
3
Both inequalities are likely to give you a range of values as solutions x and y. We need to find whether solutions x and y overlap. If the solutions overlap, we may not be able to make comparison. Otherwise, we can compare.
So, we will have to solve the given inequalities.
1. 4x - 2x ≤ 10 + 14 - 8
2x ≤ 16
x ≤ 8
2. Coverting the expression under the root,
\(16 - 8y + y^2 = (y - 4)^2\)
Solving the given inequality,
\((16 - 8y + y^2)^\frac{1}{2} ≤ 9\)
\(((y - 4)^2)^\frac{1}{2} ≤ 9\)
\(| y - 4 | ≤ 9\)
Solving this absolute value expression by distances on the number line,
-5 ≤ y ≤ 13
We see that the solutions x and y overlap.
Hence, D is the correct answer.
So, we will have to solve the given inequalities.
1. 4x - 2x ≤ 10 + 14 - 8
2x ≤ 16
x ≤ 8
2. Coverting the expression under the root,
\(16 - 8y + y^2 = (y - 4)^2\)
Solving the given inequality,
\((16 - 8y + y^2)^\frac{1}{2} ≤ 9\)
\(((y - 4)^2)^\frac{1}{2} ≤ 9\)
\(| y - 4 | ≤ 9\)
Solving this absolute value expression by distances on the number line,
-5 ≤ y ≤ 13
We see that the solutions x and y overlap.
Hence, D is the correct answer.
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Re: 4(x+2) < 2(x+5) + 14
[#permalink]
25 May 2021, 21:10
25 May 2021, 21:10
1
cbrelax wrote:
Both inequalities are likely to give you a range of values as solutions x and y. We need to find whether solutions x and y overlap. If the solutions overlap, we may not be able to make comparison. Otherwise, we can compare.
So, we will have to solve the given inequalities.
1. 4x - 2x ≤ 10 + 14 - 8
2x ≤ 16
x ≤ 8
2. Coverting the expression under the root,
\(16 - 8y + y^2 = (y - 4)^2\)
Solving the given inequality,
\((16 - 8y + y^2)^\frac{1}{2} ≤ 9\)
\(((y - 4)^2)^\frac{1}{2} ≤ 9\)
\(| y - 4 | ≤ 9\)
Solving this absolute value expression by distances on the number line,
-5 ≤ y ≤ 13
We see that the solutions x and y overlap.
Hence, D is the correct answer.
So, we will have to solve the given inequalities.
1. 4x - 2x ≤ 10 + 14 - 8
2x ≤ 16
x ≤ 8
2. Coverting the expression under the root,
\(16 - 8y + y^2 = (y - 4)^2\)
Solving the given inequality,
\((16 - 8y + y^2)^\frac{1}{2} ≤ 9\)
\(((y - 4)^2)^\frac{1}{2} ≤ 9\)
\(| y - 4 | ≤ 9\)
Solving this absolute value expression by distances on the number line,
-5 ≤ y ≤ 13
We see that the solutions x and y overlap.
Hence, D is the correct answer.
why you put modulus sign after releasing quantity from square root?
