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

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<math>\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3</math>
 
<math>\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3</math>
  
==Solutions==
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==Solution 1==
 
 
===Solution 1===
 
 
Rearranging, we find <math>3x+y=-2x+6y</math>, or <math>5x=5y\implies x=y</math>.
 
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>.
 
Substituting, we can convert the second equation into <math>\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{\textbf{(D)}\ 2}</math>.
  
===Solution 2===
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 +
 
 +
More step-by-step explanation:
 +
 
 +
<math>\frac{3x+y}{x-3y}=-2</math>
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<math>3x+y=-2\left(x-3y\right)</math>
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<math>3x+y=-2x+6y</math>
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 +
<math>5x=5y</math>
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 +
<math>x=y</math>
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<math>\frac{x+3y}{3x-y}=\frac{1+3\left(1\right)}{3\left(1\right)-1}=\frac{4}{2}=2</math>.
 +
 
 +
We choose <math>\boxed{\textbf{(D)}\ 2}</math>.
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 +
==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>
 
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>
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 +
==Solution 3==
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Let <math>y=ax</math>. The first equation converts into <math>\frac{(3+a)x}{(1-3a)x}=-2</math>, which simplifies to <math>3+a=-2(1-3a)</math>. After a bit of algebra we found out <math>a=1</math>, which means that <math>x=y</math>. Substituting <math>y=x</math> into the second equation it becomes <math>\frac{4x}{2x}=\boxed{\textbf{(D)}\ 2}</math> - mathleticguyyy
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== Solution 4 ==
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Let <math>x=1</math>. Then <math>y=1</math>. So the desired result is <math>2</math>. Select <math>\boxed{D}</math>.
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 +
~hastapasta
 +
== Video Solution ==
 +
https://youtu.be/ba6w1OhXqOQ?t=1059
 +
 +
~ pi_is_3.14
 +
 +
==Video Solution==
 +
https://youtu.be/B0NUA9011OQ
 +
 +
~savannahsolver
 +
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/zTGuz6EoBWY?t=668
 +
 +
~IceMatrix
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=3|num-a=5}}
 
{{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}}
 
{{MAA Notice}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 16:34, 13 May 2022

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 1

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


More step-by-step explanation:

$\frac{3x+y}{x-3y}=-2$

$3x+y=-2\left(x-3y\right)$

$3x+y=-2x+6y$

$5x=5y$

$x=y$

$\frac{x+3y}{3x-y}=\frac{1+3\left(1\right)}{3\left(1\right)-1}=\frac{4}{2}=2$.

We choose $\boxed{\textbf{(D)}\ 2}$.

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

Solution 4

Let $x=1$. Then $y=1$. So the desired result is $2$. Select $\boxed{D}$.

~hastapasta

Video Solution

https://youtu.be/ba6w1OhXqOQ?t=1059

~ pi_is_3.14

Video Solution

https://youtu.be/B0NUA9011OQ

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/zTGuz6EoBWY?t=668

~IceMatrix

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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