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# The system of equations above has how many solutions?

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Manager
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The system of equations above has how many solutions? [#permalink]  15 Nov 2007, 13:21
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x - y = 3
2x = 2y + 6

The system of equations above has how many solutions?

(A) None
(B) Exactly one
(C) Exactly two
(D) Exactly three
(E) Infinitely many
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Jan 2014, 02:07, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: Equations & Solutions [#permalink]  15 Nov 2007, 13:52
yogachgolf wrote:
yogachgolf wrote:
x-y = 3
2x= 2y+6

The system of equations above has how many
solutions?

(A) None
(B) Exactly one
(C) Exactly two
(D) Exactly three
(E) Infinitely many

I thought it's A as well. But OA is E?

I see your point ... A + B can be an infinite number of things... I suppose we answered it on the basis that the two solutions given gave us no CERTAIN solutions...

I think the question is a touch ambiguous, but I suppose we could have read it more closely.
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Re: Equations & Solutions [#permalink]  15 Nov 2007, 14:19
alrussell wrote:
yogachgolf wrote:
yogachgolf wrote:
x-y = 3
2x= 2y+6

The system of equations above has how many solutions?

(A) None
(B) Exactly one
(C) Exactly two
(D) Exactly three
(E) Infinitely many

I thought it's A as well. But OA is E?

I see your point ... A + B can be an infinite number of things... I suppose we answered it on the basis that the two solutions given gave us no CERTAIN solutions...

I think the question is a touch ambiguous, but I suppose we could have read it more closely.

seems ok but we cannot solve the equations do not provide any value for x and y.
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OA is E [#permalink]  15 Nov 2007, 22:12
to satisfy x-y=3

(1,-2)
(2,-1)
......
infinitely

two equation are same.

If we think these based on function, these are same linear.

Last edited by LEE SANG IL on 16 Nov 2007, 21:42, edited 1 time in total.
Manager
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Any single linear equation with more than 1 variable in it has infinite solutions (provided no constraints are given).

Edit: I do not see any ambiguity in the question.

The number of solutions for a linear equation is the number of possible values the variables can have so as to satisfy the equation. There are infinite possible values for the variables x & y in the given equation and therefore there are infinite solutions.
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Re: System of Equations [#permalink]  04 Apr 2011, 03:57
2
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petrifiedbutstanding wrote:
Attachment:
1.JPG

The system of equations above has how many solutions?
(A) None
(B) Exactly one
(C) Exactly two
(D) Exactly three
(E) Infinitely many

Sol:

$$x-y=3$$ is same as $$2x=2y+6$$

$$2x=2y+6$$
Dividing both sides by 2;
$$x=y+3$$
Subtracting y from both sides;
$$x-y=3$$

Thus, we have only one equation:
$$x-y=3$$
This has infinitely many solutions such as:
x=3,y=0
x=100,y=97
x=-100,y=-103
x=0.001, y=-2.999
x=1, y=-2
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Re: System of Equations [#permalink]  04 Apr 2011, 04:50
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Both the equations are same, so it will have infinitely many solutions.

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Re: algebra [#permalink]  06 Jul 2011, 18:53
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given two equations are exactly same.

so different values of x , will yield different values of y.

=> infinite solutions

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Re: algebra [#permalink]  07 Jul 2011, 09:47
ssarkar wrote:
x-y=3
2x=2y+6

The system of equations above has how many solutions?
(A) None
(B) Exactly one
(C) Exactly two
(D) Exactly three
(E) Infinitely many

x-y=3 ---------------1
2x=2y+6 ---------------2

Divide equation 2 by 2:
2x/2=(2y+6)/2
x=y+3
x-y=3----------------3

Equation 1 and 3 are equal and thus have infinitely many solutions:

x-y=3
x=5, y=2
x=6, y=3
x=7, y=4

Ans: "E"
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Re: algebra [#permalink]  07 Jul 2011, 20:41
1
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Expert's post
ssarkar wrote:
x-y=3
2x=2y+6

The system of equations above has how many solutions?
(A) None
(B) Exactly one
(C) Exactly two
(D) Exactly three
(E) Infinitely many

Say, the given equations are:

ax + by = c
dx + ey = f

If $$\frac{a}{d} \neq \frac{b}{e}$$, then the system of equations has a unique solution.

If $$\frac{a}{d} = \frac{b}{e} \neq \frac{c}{f}$$, then the system of equations has no solution.

$$\frac{a}{d} = \frac{b}{e} = \frac{c}{f}$$, then the system of equations has infinitely many solutions.

Here, $$\frac{1}{2} = \frac{-1}{-2} = \frac{3}{6}$$ so there are infinitely many solutions.
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Re: Pls help me to solve this problem. [#permalink]  31 Jul 2011, 04:19
tracyyahoo wrote:
x-y=3
2x=2y+6

The system of equations above has how many solutions?

a) None
b) Exactly one
c) Exactly two
d) Exactly three
e) Infinitely many

I calculate that x will substracted by the equation, and I think it is b.

Actually both the equations represent the same i.e.

2x = 2y+6
=> x=y+3
=> x-y = 3 same as eq 1

Hence there are Infinitely many solutions to the equations as there are no restrictions

Hence E
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Re: [#permalink]  05 Jan 2014, 22:36
jbs wrote:
Any single linear equation with more than 1 variable in it has infinite solutions (provided no constraints are given).

Edit: I do not see any ambiguity in the question.

The number of solutions for a linear equation is the number of possible values the variables can have so as to satisfy the equation. There are infinite possible values for the variables x & y in the given equation and therefore there are infinite solutions.

Makes proper sense
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Re: x-y = 3 2x= 2y+6 The system of equations above has how many [#permalink]  05 Jan 2014, 22:53
yogachgolf wrote:
x-y = 3
2x= 2y+6

The system of equations above has how many
solutions?

(A) None
(B) Exactly one
(C) Exactly two
(D) Exactly three
(E) Infinitely many

Equation 1 = x = y + 3
Equation 2 = x = y + 3

Since both the equations represent a single line hence there will be infinitely many solutions for this.
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Re: x-y = 3 2x= 2y+6 The system of equations above has how many   [#permalink] 05 Jan 2014, 22:53
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