Difference between revisions of "2017 AMC 10B Problems/Problem 4"

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==See Also==
 
==See Also==
 
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{{AMC12 box|year=2017|ab=B|before=First Problem|num-a=4}}
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{{AMC12 box|year=2017|ab=B|num-b=2|num-a=4}}
 
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 14:34, 20 January 2020

Problem

Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$. What is the value of $\frac{x+3y}{3x-y}$?

$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$

Solutions

Solution 1

Rearranging, we find $3x+y=-2x+6y$, or $5x=5y\implies x=y$. Substituting, we can convert the second equation into $\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{\textbf{(D)}\ 2}$.

Solution 2

Substituting each $x$ and $y$ with $1$, we see that the given equation holds true, as $\frac{3(1)+1}{1-3(1)} = -2$. Thus, $\frac{x+3y}{3x-y}=\boxed{\textbf{(D)}\ 2}$

Solution 3

Let $y=ax$. The first equation converts into $\frac{(3+a)x}{(1-3a)x}=-2$, which simplifies to $3+a=-2(1-3a)$. After a bit of algebra we found out $a=1$, which means that $x=y$. Substituting $y=x$ into the second equation it becomes $\frac{4x}{2x}=\boxed{\textbf{(D)}\ 2}$ - mathleticguyyy

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2017 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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