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

Latest revision as of 11:36, 31 March 2018

Problem

If $(x, y)$ is a solution to the system $xy=6$ and $x^2y+xy^2+x+y=63$ , 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 $x^2+y^2= \boxed{69}$

1987 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 2: / Line 2: @@
 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>,
+<math>xy=6</math> and <math>x^2y+xy^2+x+y=63</math>,
 find <math>x^2+y^2</math>.

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

Latest revision as of 11:36, 31 March 2018

Problem

Solution

See also