2017 AMC 10B Problems/Problem 4: Difference between revisions
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==Problem== | ==Problem== | ||
Supposed that <math>x</math> and <math>y</math> are nonzero real numbers such that <math>\frac{3x+y}{x-3y}=-2</math>. What is the value of <math>\frac{x+3y}{3x-y}</math>? | |||
<math>\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3</math> | |||
<math>\ | ==Solution 1== | ||
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>. | |||
{{AMC10 box|year=2017|ab= | |||
More step-by-step explanation: | |||
<math>\frac{3x+y}{x-3y}=-2</math> | |||
<math>3x+y=-2\left(x-3y\right)</math> | |||
<math>3x+y=-2x+6y</math> | |||
<math>5x=5y</math> | |||
<math>x=y</math> | |||
<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>. | |||
==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> | |||
==Solution 3== | |||
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 | |||
== 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== | |||
{{AMC10 box|year=2017|ab=B|num-b=3|num-a=5}} | |||
{{AMC12 box|year=2017|ab=B|num-b=2|num-a=4}} | |||
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Introductory Algebra Problems]] | |||
Latest revision as of 21:34, 24 August 2024
Problem
Supposed that 
and 
are nonzero real numbers such that 
. What is the value of 
?
Solution 1
Rearranging, we find 
, or 
.
Substituting, we can convert the second equation into 
.
More step-by-step explanation:
.
We choose 
.
Solution 2
Substituting each and with 
, we see that the given equation holds true, as 
. Thus, 
Solution 3
Let 
. The first equation converts into 
, which simplifies to 
. After a bit of algebra we found out 
, which means that . Substituting 
into the second equation it becomes 
- mathleticguyyy
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=1059
~ pi_is_3.14
Video Solution
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/zTGuz6EoBWY?t=668
~IceMatrix
See Also
| 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
