2014 AMC 10B Problems/Problem 9
Contents
Problem
For real numbers and , What is ?
Solution
Multiply the numerator and denominator of the LHS (left hand side) by to get . Then since and , , or choice .
Solution 2
Basic algebra at the end of the day, so simplify the numerator and the denominator. The numerator simplifies out to and the denominator simplifies out to .
This results in .
Division results in the elimination of , so we get .
is just so the equation above is .
Solving this results in .
~AkCANdo
Solution 3
Muliply both sides by to get . Then, add to both sides and subtract from both sides to get . Then, we can plug in the most simple values for z and w ( and , respectively), and find , or answer choice .
Solution 4
Let and . To find values for a and b, we can try and . However, that leaves us with a fractional solution, so scaling it by 2, we get and . Solving by adding the equations together, we get and . Now, substituting back in, we get and . Now, putting this into the desired equation with (since it will cancel out), we get . Dividing, we get .
~idk12345678
Solution 5
Set \( w = 2 \) and \( z = 1 \).
Substitute the new values into the first equation
,
,
Substitute in the second equation with new values of \( w \) and \( z \):
(2 + 1) / (2 - 1) = 3.
Answers of each equation (where X is the quotient): and
Therefore, the answers to the equations are the negatives of each other. Thus the answer is (A)
~WalkEmDownTrey
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.