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How many ordered pairs of real numbers (x,y) satisfy the following sys
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18 Mar 2019, 08:58
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26% (02:28) correct 74% (02:17) wrong based on 85 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
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Re: How many ordered pairs of real numbers (x,y) satisfy the following sys
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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.




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



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Re: How many ordered pairs of real numbers (x,y) satisfy the following sys
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18 Mar 2019, 10:36
hi longranger25 your solution is correct ..
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Re: How many ordered pairs of real numbers (x,y) satisfy the following sys
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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 ?
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Re: How many ordered pairs of real numbers (x,y) satisfy the following sys
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06 Apr 2019, 07:36
Hi Experts, chetan2u Bunuel VeritasKarishma Is there a better way or formula to find ordered pairs of such equations?
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How many ordered pairs of real numbers (x,y) satisfy the following sys
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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).



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



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



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



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






