Difference between revisions of "1988 AHSME Problems/Problem 19"
(→Solution 4) |
m (→Solution) |
||
| Line 12: | Line 12: | ||
| − | ==Solution== | + | ==Solution 1== |
| − | + | We can multiply each answer choice by <math>bx + ay</math> and then compare with the numerator. This gives <math>\boxed{\text{B}}</math>. | |
==Solution 2== | ==Solution 2== | ||
Revision as of 20:46, 16 January 2024
Problem
Simplify
Solution 1
We can multiply each answer choice by 
and then compare with the numerator. This gives 
.
Solution 2
Expanding everything in the brackets, we get
. We can then group numbers up in pairs so they equal 
:
We get .
-ThisUsernameIsTaken
Solution 3
If you were out of time and your algebra isn't that good, you could just plug in some values for the variables and see which answer choice works.
Solution 4 (fastest)
After regrouping, the numerator becomes 
. Factoring further, we get 
. After dividing, we get 
, which can be factored as 
, so the answer is .
See also
| 1988 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
