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Given x^3 + y^3 = 19 and (x + y)(8 + xy) = 2. What is the sum of x and
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Updated on: 03 Oct 2021, 02:21
Updated on: 03 Oct 2021, 02:21
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Given \(x^3 + y^3 = 19\) and \((x + y)(8 + xy) = 2\). What is the sum of x and y?
A. 1
B. -6
C. -2
D. 3
E. 6
A. 1
B. -6
C. -2
D. 3
E. 6
Originally posted by huongguyen on 03 Oct 2021, 01:57.
Last edited by Bunuel on 03 Oct 2021, 02:21, edited 1 time in total.
Last edited by Bunuel on 03 Oct 2021, 02:21, edited 1 time in total.
Renamed the topic, edited the question and edited the tags.
Re: Given x^3 + y^3 = 19 and (x + y)(8 + xy) = 2. What is the sum of x and
[#permalink]
03 Oct 2021, 11:04
03 Oct 2021, 11:04
3
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Expert Reply
huongguyen wrote:
Given \(x^3 + y^3 = 19\) and \((x + y)(8 + xy) = 2\). What is the sum of x and y?
A. 1
B. -6
C. -2
D. 3
E. 6
A. 1
B. -6
C. -2
D. 3
E. 6
The first thing that I would tackle is the equation \(x^3+y^3=19\)
Yes, x and y could be fractions but the two equations dealing with integers does give an indication that the values may be integers themselves.
Now 0, 1, 8, 27, 64, 125 are cubes.
Can we get 19 from any two of them?
Yes, 27-8=19
Thus one of the variables could be 3 and other -2.
Let us check the other equation too.
(3+(-2))(8+3*(-2))=(3-2)(8-6)=2
Yes , fits in perfectly.
So x +y =1
A
Given x^3 + y^3 = 19 and (x + y)(8 + xy) = 2. What is the sum of x and
[#permalink]
05 Oct 2021, 07:36
05 Oct 2021, 07:36
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\(x^3+y^3 = 19\)
OR\( (x+y) ( x^2 - xy +y^2) = 19 * 1\) OR \(1* 19\)
Hence,\( x+y = 1\) OR 19[/m] --------------- (1)
\((x+y)(8+xy)=2\) OR \(2*1 \) OR \(1*2\)
Hence, \(x+y = 1\) OR 2[/m] ----------------- (2)
With both equations \(x+y = 1\)
Good that question has 19 and 2 : prime numbers
OR\( (x+y) ( x^2 - xy +y^2) = 19 * 1\) OR \(1* 19\)
Hence,\( x+y = 1\) OR 19[/m] --------------- (1)
\((x+y)(8+xy)=2\) OR \(2*1 \) OR \(1*2\)
Hence, \(x+y = 1\) OR 2[/m] ----------------- (2)
With both equations \(x+y = 1\)
Good that question has 19 and 2 : prime numbers
