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

Revision as of 10:10, 26 July 2017

Problem

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

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

Solution

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

@@ Line 7: / Line 7: @@
 Rearranging, we find <math>3x+y=-2x+6y</math>, or <math>5x=5y\implies x=y</math>
 Substituting, we can convert the second equation into <math>\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{\textbf{(D)}\ 2}</math>
-==Solution 2==
-Substituting each <math>x</math> and <math>y</math> with <math>1</math>, we see that the given equation holds true, as <math>\frac{3(1)+1}{1-3(1)} = -2</math>. Thus, <math>\frac{x+3y}{3x-y}=\boxed{\textbf{(D)}\ 2}</math>
-{{AMC10 box|year=2017|ab=B|num-b=3|num-a=5}}
-{{MAA Notice}}

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

Revision as of 10:10, 26 July 2017

Problem

Solution