Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1988 AHSME Problems/Problem 19"

(→‎Solution 4)
(→‎Solution 4 (fastest))
 
(3 intermediate revisions by the same user not shown)
Line 12: Line 12:
  
  
−
==Solution==
+
==Solution 1==
−
The fastest way is to multiply each answer choice by <math>bx + ay</math> and then compare to the numerator. This gives <math>\boxed{\text{B}}</math>.
+
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==
Line 34: Line 34:
 
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.
 
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==
+
==Solution 4 (fastest)==
−
After regrouping, the numerator becomes <math>(bx+ay)(a^2x^2+b^2y^2)+2bxa^2y^2+2ayb^2x^2</math>. Factoring further, we get <math>(bx+ay)(a^2x^2+b^2y^2)+2bxay(bx+ay)</math>. After dividing, we get <math>a^2x^2+b^2y^2+2bxay</math>, which can be factored as <math>(ax+by)^2</math>, so the answer is B.
+
After regrouping, the numerator becomes <math>(bx+ay)(a^2x^2+b^2y^2)+2bxa^2y^2+2ayb^2x^2</math>. Factoring further, we get <math>(bx+ay)(a^2x^2+b^2y^2)+2bxay(bx+ay)</math>. After dividing, we get <math>a^2x^2+b^2y^2+2bxay</math>, which can be factored as <math>(ax+by)^2</math>, so the answer is <math>\boxed{\text{B}}</math>.
 +
 
 +
-Pengu14
  
 
== See also ==
 
== See also ==

Latest revision as of 20:47, 16 January 2024

Problem

Simplify

$\frac{bx(a^2x^2 + 2a^2y^2 + b^2y^2) + ay(a^2x^2 + 2b^2x^2 + b^2y^2)}{bx + ay}$

$\textbf{(A)}\ a^2x^2 + b^2y^2\qquad \textbf{(B)}\ (ax + by)^2\qquad \textbf{(C)}\ (ax + by)(bx + ay)\qquad\\ \textbf{(D)}\ 2(a^2x^2+b^2y^2)\qquad \textbf{(E)}\ (bx+ay)^2$


Solution 1

We can multiply each answer choice by $bx + ay$ and then compare with the numerator. This gives $\boxed{\text{B}}$.

Solution 2

Expanding everything in the brackets, we get $\frac{ba^2x^3 + 2ba^2xy^2 + b^3xy^2 + a^3x^2y + 2ab^2x^2y + ab^2y^3}{bx + ay}$. We can then group numbers up in pairs so they equal $n(bx + ay)$:

$= \frac{ba^2x^3 + a^3x^2y + 2ab^2x^2y + 2ba^2xy^2 + b^3xy^2 + ab^2y^3}{bx+ay}$

$= \frac{bx + ay(a^2x^2) + bx + ay(2baxy) + bx + ay(b^2y^2)}{bx+ay}$

$= a^2x^2 + 2baxy + b^2y^2$

$= (ax + by)^2$

We get $\boxed{\text{B}}$.

-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 $(bx+ay)(a^2x^2+b^2y^2)+2bxa^2y^2+2ayb^2x^2$. Factoring further, we get $(bx+ay)(a^2x^2+b^2y^2)+2bxay(bx+ay)$. After dividing, we get $a^2x^2+b^2y^2+2bxay$, which can be factored as $(ax+by)^2$, so the answer is $\boxed{\text{B}}$.

-Pengu14

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1988_AHSME_Problems/Problem_19&oldid=210972"