Difference between revisions of "2017 AMC 10B Problems/Problem 4"
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(removed solution 4, same as solution 2.) |
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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> | ||
| − | + | ==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 | + | |
| + | |||
| + | 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> | 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== | ==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| | + | {{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 21:34, 24 August 2024
Contents
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
