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The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions.

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The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions.  [#permalink]

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New post 11 Jul 2017, 03:36
1
3
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

44% (02:10) correct 56% (02:36) wrong based on 71 sessions

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The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions.  [#permalink]

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New post 11 Jul 2017, 03:51
Bunuel wrote:
The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions. What is the value of (a + b)?

(1) c = 3
(2) b = a + 1


As given condition, we conclude that those two equations are identical.

Hence \(\frac{a}{b}=\frac{2}{c}=\frac{6}{9}=\frac{2}{3}\)

\(\frac{2}{c} = \frac{2}{3} \implies c=3\)

(1) \(c=3\). No information about \(a\) and \(b\). Insufficient.

(2) \(b=a+1 \implies \frac{a}{b}= \frac{b-1}{b}=1-\frac{1}{b}=\frac{2}{3} \)
\(\implies \frac{1}{b}=\frac{1}{3} \implies b=3 \implies a=2 \implies a+b=5\).
Sufficient.

The answer is B
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Re: The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions.  [#permalink]

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New post 11 Jul 2017, 07:53
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nguyendinhtuong wrote:
(2) \(b=a+1 \implies \frac{a}{b}= \frac{b-1}{b}=1-\frac{1}{b}=\frac{2}{3} \)
\(\implies \frac{1}{b}=\frac{1}{3} \implies b=3 \implies a=4 \implies a+b=7\).


Great solution, but here, to find a, you added 1 to b, instead of subtracting 1 from b (the solution should be a=2 and b=3).

I find the wording of the question very problematic. When I first read the question:

"The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions"

I think 'of course they do, each is one equation with three or four unknowns'. They mean there are infinite solutions when the two equations are solved simultaneously. And they further mean there are infinite solutions specifically for x and y, not for a, b or c. The question needs to say that; someone who has studied a lot of algebra might assume that, because it's a convention that x and y are your 'variables' and early letters in the alphabet are constants, but there's no logical reason that needs to be true.
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Re: The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions.  [#permalink]

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New post 11 Jul 2017, 08:27
Bunuel wrote:
The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions. What is the value of (a + b)?

(1) c = 3
(2) b = a + 1


I think answer should be D.
From 1. c=3 so bx+3y = 9 and ax+2y = 6. Now to have these 2 equations infinite solution b=3 and a=2. Then both these equations wil become x+y = 3.
From 2. Replace b = a+1 in a/b = 2/3. it gives a = 2 and b =3. Hence a+b = 5.
Either of these is sufficient. Hence D.
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Re: The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions.  [#permalink]

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New post 11 Jul 2017, 09:19
IanStewart wrote:
nguyendinhtuong wrote:
(2) \(b=a+1 \implies \frac{a}{b}= \frac{b-1}{b}=1-\frac{1}{b}=\frac{2}{3} \)
\(\implies \frac{1}{b}=\frac{1}{3} \implies b=3 \implies a=4 \implies a+b=7\).


Great solution, but here, to find a, you added 1 to b, instead of subtracting 1 from b (the solution should be a=2 and b=3).

I find the wording of the question very problematic. When I first read the question:

"The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions"

I think 'of course they do, each is one equation with three or four unknowns'. They mean there are infinite solutions when the two equations are solved simultaneously. And they further mean there are infinite solutions specifically for x and y, not for a, b or c. The question needs to say that; someone who has studied a lot of algebra might assume that, because it's a convention that x and y are your 'variables' and early letters in the alphabet are constants, but there's no logical reason that needs to be true.

Lol, thank you for pointing out my mistake :-D

I must agree with you :-D "equations" must be replaced as "simultaneous equations". However, I still treated them as simultaneous equations, not two different equations :-D kudos :lol:

Posted from my mobile device
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Re: The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions.  [#permalink]

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New post 11 Jul 2017, 11:02
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[b]The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions. What is the value of (a + b)?

(1) c = 3
(2) b = a + 1

given Line 1 & 2 got infinite solution ==> implies both the lines overlap each other ==> which implies both the lines are same

\(\frac{x}{6/a}\) + \(\frac{y}{3}\) = 1

\(\frac{x}{9/b}\) + \(\frac{x}{9/c}\) = 1


so the intercept of the lines must be same as well

\(\frac{6}{a}\) = \(\frac{9}{b}\) & \(\frac{9}{c}\) = 3

so we have ==>c = 3 and \(\frac{a}{b}\) = \(\frac{2}{3}\)

statement 1 gives us the value which we already know : so no use

statement 2: b = a + 1 so \(\frac{a}{a + 1}\) = \(\frac{2}{3}\)
a = 2 & b = 3
a + b = 5

Hence Option B is the answer

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Re: The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions.  [#permalink]

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New post 14 Jul 2017, 11:46
ax + 2y = 6
bx + cy = 9
have infinte solutions, means that both equations are identical
his implies that a/b = 6/9 = 2/3
or b = 1.5a
so we have to find value of 2.5a

Statement 1- gives us value of c which 3/2*2 - no extra information
hence insufficient

Statement 2 - gives us another relation between a and b
if b = a+1 and b = 1.5 a
means a = 0.5
hence a+b = 1.25
sufficient

answer is B

+1 kudos if you like the post.
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Re: The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions.  [#permalink]

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New post 15 Jul 2017, 02:57
sonikavadhera wrote:
ax + 2y = 6
bx + cy = 9
have infinte solutions, means that both equations are identical
his implies that a/b = 6/9 = 2/3
or b = 1.5a
so we have to find value of 2.5a

Statement 1- gives us value of c which 3/2*2 - no extra information
hence insufficient

Statement 2 - gives us another relation between a and b
if b = a+1 and b = 1.5 a
means a = 0.5
hence a+b = 1.25

sufficient

answer is B

+1 kudos if you like the post.


Thank you for the solution but a minor mistake (no impact on answer)
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Re: The equations ax + 2y = 6 and bx + cy = 9 have infinite solutions.   [#permalink] 15 Jul 2017, 02:57
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