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Given the two equations 3r + s = 17 and r + 2s = 9, by how much does

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Math Expert
Joined: 02 Sep 2009
Posts: 49231
Given the two equations 3r + s = 17 and r + 2s = 9, by how much does  [#permalink]

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07 Feb 2016, 10:04
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5% (low)

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89% (01:10) correct 11% (01:02) wrong based on 70 sessions

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Given the two equations 3r + s = 17 and r + 2s = 9, by how much does r exceed s?

A. 3
B. 4
C. 5
D. 6
E. 7

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Joined: 18 Aug 2013
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Re: Given the two equations 3r + s = 17 and r + 2s = 9, by how much does  [#permalink]

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13 Feb 2016, 09:56
Solve by Simultaneous Equations.

r=5
s=2
Intern
Joined: 24 Jul 2017
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Re: Given the two equations 3r + s = 17 and r + 2s = 9, by how much does  [#permalink]

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26 Sep 2017, 02:42
Bunuel wrote:
Given the two equations 3r + s = 17 and r + 2s = 9, by how much does r exceed s?

A. 3
B. 4
C. 5
D. 6
E. 7

solve for these 2 eqns , you ll get r =5 & s =2
hence r-s = 5-2 = 3
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Re: Given the two equations 3r + s = 17 and r + 2s = 9, by how much does  [#permalink]

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29 Sep 2017, 10:32
Bunuel wrote:
Given the two equations 3r + s = 17 and r + 2s = 9, by how much does r exceed s?

A. 3
B. 4
C. 5
D. 6
E. 7

We can multiply the second equation by -3 and we have:

-3r - 6s = -27

When we add this equation to the first equation, the terms containing r cancel out, and we now have:

-5s = -10

s = 2

Thus:

r + 2(2) = 9

r = 5

So, r is 3 greater than s.

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Re: Given the two equations 3r + s = 17 and r + 2s = 9, by how much does &nbs [#permalink] 29 Sep 2017, 10:32
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