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

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==Solution==
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==Solution 1==
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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>.
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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==
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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.
  
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==Solution 4==
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==Solution 4 (fastest)==
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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 C.
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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>.
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-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

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