Aug 20 08:00 PM PDT  09:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Aug 20 09:00 PM PDT  10:00 PM PDT Take 20% off the plan of your choice, now through midnight on Tuesday, 8/20 Aug 22 09:00 PM PDT  10:00 PM PDT What you'll gain: Strategies and techniques for approaching featured GMAT topics, and much more. Thursday, August 22nd at 9 PM EDT Aug 24 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC Aug 25 09:00 AM PDT  12:00 PM PDT Join a FREE 1day verbal workshop and learn how to ace the Verbal section with the best tips and strategies. Limited for the first 99 registrants. Register today! Aug 25 08:00 PM PDT  11:00 PM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE.
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Status: Manager
Joined: 02 Nov 2018
Posts: 283
Location: Bangladesh

How many ordered pairs of real numbers (x,y) satisfy the following sys
[#permalink]
Show Tags
18 Mar 2019, 08:58
Question Stats:
24% (02:30) correct 76% (02:25) wrong based on 116 sessions
HideShow timer Statistics
How many ordered pairs of real numbers\((x,y)\) satisfy the following system of equations? \(x + 3y = 3\) \(x − y\)= 1 (A) 1 (B) 2 (C) 3 (D) 4 (E) 8
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Give a kudos if u find my post helpful. kudos motivates active discussions




Intern
Joined: 02 Mar 2019
Posts: 3

Re: How many ordered pairs of real numbers (x,y) satisfy the following sys
[#permalink]
Show Tags
08 Apr 2019, 11:18
Detailed Solution:
1. Solve two equations (Solving by Combination:
x+3y=3 x−y= 1
Here for the second equation, give values for x and y
Case 1: x=+x and y=+y x+3y=3 xy = 1 Solve the equation x=1.5 and y=0.5
Case 2: x=x and y=y x+3y=3 x+y = 1 Solve the equation x=0 and y=1
Case 3: x=x and y=y x+3y=3 xy = 1 Solve the equation x=3 and y=2
Case 4: x=+x and y=y Try yourself, you will get one of the same solution above
Hence we have three solution pairs > (x,y) = (1.5,0.5), (0,1) and (3,2)
HENCE SOLVED.




Intern
Joined: 24 Jul 2013
Posts: 29

Re: How many ordered pairs of real numbers (x,y) satisfy the following sys
[#permalink]
Show Tags
18 Mar 2019, 10:20
Noshad wrote: How many ordered pairs of real numbers\((x,y)\) satisfy the following system of equations?
\(x + 3y = 3\)
\(x − y\)= 1
(A) 1
(B) 2
(C) 3
(D) 4
(E) 8 Is it C? I am getting 3 points (3,2), (0,1) and (1.5,0,5)



Senior Manager
Status: Manager
Joined: 02 Nov 2018
Posts: 283
Location: Bangladesh

Re: How many ordered pairs of real numbers (x,y) satisfy the following sys
[#permalink]
Show Tags
18 Mar 2019, 10:36
hi longranger25 your solution is correct ..
_________________
Give a kudos if u find my post helpful. kudos motivates active discussions



Director
Joined: 27 May 2012
Posts: 838

Re: How many ordered pairs of real numbers (x,y) satisfy the following sys
[#permalink]
Show Tags
21 Mar 2019, 13:55
longranger25 wrote: Noshad wrote: How many ordered pairs of real numbers\((x,y)\) satisfy the following system of equations?
\(x + 3y = 3\)
\(x − y\)= 1
(A) 1
(B) 2
(C) 3
(D) 4
(E) 8 Is it C? I am getting 3 points (3,2), (0,1) and (1.5,0,5) Please can you show your working ?
_________________



Manager
Joined: 19 Sep 2017
Posts: 179
Location: United Kingdom
GPA: 3.9
WE: Account Management (Other)

Re: How many ordered pairs of real numbers (x,y) satisfy the following sys
[#permalink]
Show Tags
06 Apr 2019, 07:36
Hi Experts, chetan2u Bunuel VeritasKarishma Is there a better way or formula to find ordered pairs of such equations?
_________________
Cheers!!
~Whatever one practices, becomes a habit. Make sure you practice the right ones.~



Intern
Joined: 07 Apr 2019
Posts: 3

How many ordered pairs of real numbers (x,y) satisfy the following sys
[#permalink]
Show Tags
08 Apr 2019, 17:36
Noshad wrote: How many ordered pairs of real numbers\((x,y)\) satisfy the following system of equations?
\(\begin{align}x + 3y &= 3 \quad (1)\\ x − y &= 1\quad (2)\end{align}\)
(A) 1 (B) 2 (C) 3 (D) 4 (E) 8 One can drop an absolute value by introducing cases, e.g. \(\lvert x 7\rvert = \begin{cases}\phantom{(} x 7&,\text{if } x \geq 7 \\ (x 7) &, \text{if } 7 > x\,. \end{cases}\) Depending on whether the argument is positive or negative (or zero), we have to apply a change of sign. Dropping all three absolute values in equation (2) results in expressions of the form \(\pm (\pm x \pm y) = 1\) (all combinations possible), which can be written as cases \(x = \begin{cases}\phantom{} y +1 &, C_1(x,y) \\ \phantom{} y 1 &, C_2(x,y) \\ y +1 &, C_3(x,y) \\ y 1 &, C_4(x,y) \end{cases}\) where I should (but don't) keep track of the conditions imposed on \(x\) and \(y\) in the four cases. The cases represent straight (half) lines in the plane, none of which is parallel to the line of equation (1). Therefore, each of the four lines has exactly one point in common with (1). The points turn out to be \((3,2), (0,1), (1.5,0.5)\). Since I didn't keep track of the conditions \(C_k\), I have to check if the above pairs solve equation (2), they do. The correct answer is (C).



Intern
Joined: 05 Mar 2018
Posts: 29

Re: How many ordered pairs of real numbers (x,y) satisfy the following sys
[#permalink]
Show Tags
09 Apr 2019, 08:15
UranousCold wrote: Detailed Solution:
1. Solve two equations (Solving by Combination:
x+3y=3 x−y= 1
Here for the second equation, give values for x and y
Case 1: x=+x and y=+y x+3y=3 xy = 1 Solve the equation x=1.5 and y=0.5
Case 2: x=x and y=y x+3y=3 x+y = 1 Solve the equation x=0 and y=1
Case 3: x=x and y=y x+3y=3 xy = 1 Solve the equation x=3 and y=2
Case 4: x=+x and y=y Try yourself, you will get one of the same solution above
Hence we have three solution pairs > (x,y) = (1.5,0.5), (0,1) and (3,2)
HENCE SOLVED. Why have we not considered the outer modulus while solving? x and y may be +ve or ve, but what about the outermost mod? considering that would make the solution lengthier and more time consuming.



Intern
Joined: 02 Mar 2019
Posts: 3

Re: How many ordered pairs of real numbers (x,y) satisfy the following sys
[#permalink]
Show Tags
09 Apr 2019, 19:27
MeBossBaby : Thanks for the question. Absolute value equations will generally have two solutions. For Example: x could be +2 or 2 as the absolute value of x here simply refers how far number +2 or 2 is away from zero in the number line. This means we have two solutions, positive and negative. Even If I say x with double absolute value x and apply the two values +2 and 2, I will get the same solution. +2 = +2 and 2 = +2, this mean x can have still hold two values +2 and 2. We can apply this same principle on the two variable equation x−y= 1, and can safely say it can have four solutions +x, x, +y and y.



Intern
Joined: 02 Mar 2019
Posts: 3

Re: How many ordered pairs of real numbers (x,y) satisfy the following sys
[#permalink]
Show Tags
09 Apr 2019, 20:14
MeBossBaby wrote: UranousCold wrote: Detailed Solution:
1. Solve two equations (Solving by Combination:
x+3y=3 x−y= 1
Here for the second equation, give values for x and y
Case 1: x=+x and y=+y x+3y=3 xy = 1 Solve the equation x=1.5 and y=0.5
Case 2: x=x and y=y x+3y=3 x+y = 1 Solve the equation x=0 and y=1
Case 3: x=x and y=y x+3y=3 xy = 1 Solve the equation x=3 and y=2
Case 4: x=+x and y=y Try yourself, you will get one of the same solution above
Hence we have three solution pairs > (x,y) = (1.5,0.5), (0,1) and (3,2)
HENCE SOLVED. Why have we not considered the outer modulus while solving? x and y may be +ve or ve, but what about the outermost mod? considering that would make the solution lengthier and more time consuming. MeBossBaby : Thanks for the question. Absolute value equations will generally have two solutions. For Example: x could be +2 or 2 as the absolute value of x here simply refers how far number +2 or 2 is away from zero in the number line. This means we have two solutions, positive and negative. Even If I say x with double absolute value x and apply the two values +2 and 2, I will get the same solution. +2 = +2 and 2 = +2, this mean x can have still hold two values +2 and 2. We can apply this same principle on the two variable equation x−y= 1, and can safely say it can have four solutions +x, x, +y and y.



Intern
Joined: 05 Mar 2018
Posts: 29

Re: How many ordered pairs of real numbers (x,y) satisfy the following sys
[#permalink]
Show Tags
11 Apr 2019, 00:32
Okay.... I think I get it now.. Thanks UranousCold!




Re: How many ordered pairs of real numbers (x,y) satisfy the following sys
[#permalink]
11 Apr 2019, 00:32






