Difference between revisions of "1987 AHSME Problems/Problem 15"

(Created page with "==Problem== If <math>(x, y)</math> is a solution to the system <math>xy=6 \qquad \text{and} \qquad x^2y+xy^2+x+y=63</math>, find <math>x^2+y^2</math>. <math>\textbf{(A)}\ 13 \q...")
 
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\textbf{(E)}\ 81    </math>
 
\textbf{(E)}\ 81    </math>
  
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==Solution==
  
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First note that <math>x^2y+xy^2+x+y= (xy+1)(x+y)</math>. Substituting <math>6</math> for <math>xy</math> gives <math>7(x+y)= 63</math>, giving a result of <math>x+y=9</math>. Squaring this equation and subtracting by <math>12</math>, gives us <math>x^2+y^2= \boxed{69}</math>
  
 
== See also ==
 
== See also ==

Revision as of 23:05, 26 August 2016

Problem

If $(x, y)$ is a solution to the system $xy=6 \qquad \text{and} \qquad x^2y+xy^2+x+y=63$, find $x^2+y^2$.

$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ \frac{1173}{32} \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 69 \qquad \textbf{(E)}\ 81$

Solution

First note that $x^2y+xy^2+x+y= (xy+1)(x+y)$. Substituting $6$ for $xy$ gives $7(x+y)= 63$, giving a result of $x+y=9$. Squaring this equation and subtracting by $12$, gives us $x^2+y^2= \boxed{69}$

See also

1987 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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