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

(Created page with "==Problem== Simplify <math>\frac{bx(a^2x^2 + 2a^2y^2 + b^2y^2) + ay(a^2x^2 + 2b^2x^2 + b^2y^2)}{bx + ay}</math> <math>\textbf{(A)}\ a^2x^2 + b^2y^2\qquad \textbf{(B)}\ (ax + b...")
 
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==Solution==
 
==Solution==
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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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==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.
  
 
== See also ==
 
== See also ==

Revision as of 08:17, 23 September 2021

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

The fastest way is to multiply each answer choice by $bx + ay$ and then compare to 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.

See also

1988 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AHSME Problems and Solutions

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