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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # How many ordered pairs of real numbers (x,y) satisfy the following sys

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Senior Manager  P
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Joined: 02 Nov 2018
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How many ordered pairs of real numbers (x,y) satisfy the following sys  [#permalink]

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16 00:00

Difficulty:   95% (hard)

Question Stats: 23% (02:27) correct 77% (02:22) wrong based on 168 sessions

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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
Intern  B
Joined: 02 Mar 2019
Posts: 3
Re: How many ordered pairs of real numbers (x,y) satisfy the following sys  [#permalink]

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6
2
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
x-y = 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
-x-y = 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.
##### General Discussion
Intern  B
Joined: 24 Jul 2013
Posts: 28
Re: How many ordered pairs of real numbers (x,y) satisfy the following sys  [#permalink]

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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  P
Status: Manager
Joined: 02 Nov 2018
Posts: 281
Re: How many ordered pairs of real numbers (x,y) satisfy the following sys  [#permalink]

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1
hi longranger25
VP  V
Joined: 27 May 2012
Posts: 1027
Re: How many ordered pairs of real numbers (x,y) satisfy the following sys  [#permalink]

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1
longranger25 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)

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Posts: 248
Location: United Kingdom
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Re: How many ordered pairs of real numbers (x,y) satisfy the following sys  [#permalink]

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Hi Experts,
chetan2u Bunuel VeritasKarishma Is there a better way or formula to find ordered pairs of such equations?
_________________
Cheers!!
Intern  B
Joined: 07 Apr 2019
Posts: 3
How many ordered pairs of real numbers (x,y) satisfy the following sys  [#permalink]

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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  B
Joined: 05 Mar 2018
Posts: 37
Re: How many ordered pairs of real numbers (x,y) satisfy the following sys  [#permalink]

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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
x-y = 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
-x-y = 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  B
Joined: 02 Mar 2019
Posts: 3
Re: How many ordered pairs of real numbers (x,y) satisfy the following sys  [#permalink]

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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  B
Joined: 02 Mar 2019
Posts: 3
Re: How many ordered pairs of real numbers (x,y) satisfy the following sys  [#permalink]

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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
x-y = 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
-x-y = 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  B
Joined: 05 Mar 2018
Posts: 37
Re: How many ordered pairs of real numbers (x,y) satisfy the following sys  [#permalink]

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1
Okay.... I think I get it now..
Thanks UranousCold! Re: How many ordered pairs of real numbers (x,y) satisfy the following sys   [#permalink] 10 Apr 2019, 23:32

# How many ordered pairs of real numbers (x,y) satisfy the following sys  