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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>.
  
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==Solution 2==
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Expanding everything in the brackets, we get
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<math>\frac{ba^2x^3 + 2ba^2xy^2 + b^3xy^2 + a^3x^2y + 2ab^2x^2y + ab^2y^3}{bx + ay}</math>. We can then group numbers up in pairs so they equal <math>n(bx + ay)</math>:
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<math>= \frac{ba^2x^3 + a^3x^2y + 2ab^2x^2y + 2ba^2xy^2 + b^3xy^2 + ab^2y^3}{bx+ay}</math>
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<math>= \frac{bx + ay(a^2x^2) + bx + ay(2baxy) + bx + ay(b^2y^2)}{bx+ay}</math>
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<math>= a^2x^2 + 2baxy + b^2y^2</math>
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<math>= (ax + by)^2</math>
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We get <math>\boxed{\text{B}}</math>.
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-ThisUsernameIsTaken
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==Solution 3==
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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.
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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 <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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