Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
06 Dec 2021, 09:15
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
25% (02:07) correct
75%
(02:28)
wrong
based on 93
sessions
History
Date
Time
Result
Not Attempted Yet
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
Are You Up For the Challenge: 700 Level Questions
\(x + 3y = 3\)
\(||x| − |y||\)= 1
(A) 1
(B) 2
(C) 3
(D) 4
(E) 8
Are You Up For the Challenge: 700 Level Questions
07 Dec 2021, 00:32
Kudos
Bookmarks
Bunuel
This means that the difference between |x| and |y| is 1.
Say, x>0 and |x|=2, then y can be 3, -3, 1 or -1.
If x<0 and |x|=2, then y can again be 3, -3, 1 or -1.
\(||x| − |y||\)= 1 will give us four cases.
1) Both x and y are non negative.
\(|x-y|\)= 1
a) x>y........x-y = 1 and x+3y = 3
x=1+y and x=3-3y.............=> \(1+y=3-3y\) or \(4y=2\) or \(y=\frac{1}{2}\) and \(x = \frac{3}{2}\)........Possible
b) x<y........y-x = 1 and x+3y = 3 .............
x=y-1 and x=3-3y.............=> \(y-1=3-3y\) or \(4y=4\) or \(y=1\) and \(x = 0\)........Possible
2) Both x and y are negative.
\(|-x-(-y)|\)= 1 or |y-x|=1
But x+3y will become negative, so NOT possible.
3) y is positive and x is negative.
\(|-x-y|\)= 1
a) x+y = 1 and x+3y = 3
x=1-y and x=3-3y.............=> \(1-y=3-3y\) or \(2y=2\) or \(y=1\) and \(x = 0\)........ Not Possible
b) -(x+y) = 1 and x+3y = 3
x=-1-y and x=3-3y.............=> \(-1-y=3-3y\) or \(2y=4\) or \(y=2\) and \(x = -3\)........Possible
4) x is positive and y is negative.
\(|x-(-y)|\)= 1 or |x+y|=1
a) x+y = 1 and x+3y = 3
x=1-y and x=3-3y.............=> \(1-y=3-3y\) or \(2y=2\) or \(y=1\) and \(x = 0\)........ Not Possible
b) -(x+y) = 1 and x+3y = 3
x=-1-y and x=3-3y.............=> \(-1-y=3-3y\) or \(2y=4\) or \(y=2\) and \(x = -3\)........ Not Possible
Possible cases of (x,y) = (3/2,1/2); (1,0); and (-3,2) => 3 cases
C
17 Feb 2024, 23:17
Kudos
Bookmarks
From the first equation, x + 3y = 3, we can rewrite it as x = 3 - 3y.
Substitute the expression for x from step 1 into the second equation: |3 - 3y| - |y| = 1.
Since the absolute value equations can have different cases, we need to consider all possibilities:
a. When 3 - 3y ≥ 0 and y ≥ 0:
3 - 3y - y = 1
-4y = -2
y = -0.5
Substituting y = -0.5 back into x = 3 - 3y gives x = 4.5
So one solution is (4.5, -0.5).
b. When 3 - 3y ≥ 0 and y < 0:
3 - 3y + y = 1
-2y = -2
y = 1
Substituting y = 1 back into x = 3 - 3y gives x = -2
So another solution is (-2, 1).
c. When 3 - 3y < 0 and y ≥ 0:
-(3 - 3y) - y = 1
-3 + 3y - y = 1
2y = 4
y = 2
Substituting y = 2 back into x = 3 - 3y gives x = -3
So another solution is (-3, 2).
Therefore, there are 3 ordered pairs of real numbers (x,y) that satisfy the given system of equations. The correct answer is (C) 3.
Substitute the expression for x from step 1 into the second equation: |3 - 3y| - |y| = 1.
Since the absolute value equations can have different cases, we need to consider all possibilities:
a. When 3 - 3y ≥ 0 and y ≥ 0:
3 - 3y - y = 1
-4y = -2
y = -0.5
Substituting y = -0.5 back into x = 3 - 3y gives x = 4.5
So one solution is (4.5, -0.5).
b. When 3 - 3y ≥ 0 and y < 0:
3 - 3y + y = 1
-2y = -2
y = 1
Substituting y = 1 back into x = 3 - 3y gives x = -2
So another solution is (-2, 1).
c. When 3 - 3y < 0 and y ≥ 0:
-(3 - 3y) - y = 1
-3 + 3y - y = 1
2y = 4
y = 2
Substituting y = 2 back into x = 3 - 3y gives x = -3
So another solution is (-3, 2).
Therefore, there are 3 ordered pairs of real numbers (x,y) that satisfy the given system of equations. The correct answer is (C) 3.
