2017 AMC 10B Problems/Problem 4: Difference between revisions

Revision as of 13:34, 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

2017 AMC 10B (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

@@ Line 18: / Line 18: @@
 ==See Also==
 {{AMC10 box|year=2017|ab=B|num-b=3|num-a=5}}
-{{AMC12 box|year=2017|ab=B|before=First Problem|num-a=4}}
+{{AMC12 box|year=2017|ab=B|num-b=2|num-a=4}}
 {{MAA Notice}}
 [[Category:Introductory Algebra Problems]]

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