Difference between revisions of "2017 AMC 10B Problems/Problem 4"
m |
|||
Line 15: | Line 15: | ||
===Solution 3=== | ===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 | 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/B0NUA9011OQ | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Revision as of 12:03, 18 June 2020
Contents
[hide]Problem
Supposed that and
are nonzero real numbers such that
. What is the value of
?
Solutions
Solution 1
Rearranging, we find , or
.
Substituting, we can convert the second equation into
.
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
~savannahsolver
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.