Difference between revisions of "2017 AMC 10B Problems/Problem 4"
| Line 14: | Line 14: | ||
===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> | + | 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 that <math>a=1</math>, which means <math>x=y</math>. Substituting <math>x=y</math> into the second equation it becomes <math>\frac{4x}{2x}=\boxed{\textbf{(D)}\ 2}</math> - mathleticguyyy |
==See Also== | ==See Also== | ||
Revision as of 19:41, 15 September 2018
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 that 
, which means 
. Substituting into the second equation it becomes 
- mathleticguyyy
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 First Problem |
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. 
