Difference between revisions of "2000 AMC 10 Problems/Problem 15"

(Problem)
m (Solution)
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
<math>ab=a-b</math>
+
<math>\begin{align*}
 +
\frac{a}{b}+\frac{b}{a}-ab &= \frac{a^2+b^2}{ab}-ab \\
 +
&= \frac{-a^2b^2+a^2+b^2}{ab}\\
 +
\end(align*}</math>
  
<math>\frac{a}{b}+\frac{b}{a}-ab=\frac{a^2+b^2}{ab}-ab=\frac{-a^2b^2+a^2+b^2}{ab}</math>
+
Substituting <math>ab=a-b</math>, we get
  
<math>\frac{-a^2+2ab-b^2+a^2+b^2}{ab}=2</math>.
+
<math>\begin{align*}
 +
\frac{-(a+b)^2+a^2+b^2}{ab} &= \frac{-a^2+2ab-b^2+a^2+b^2}{ab} \\
 +
&= \frac{2ab}{ab} \\
 +
&= 2 \\
 +
\end{align*}</math>
  
E.
+
<math>\boxed{\text{E}}</math>
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2000|num-b=14|num-a=16}}
 
{{AMC10 box|year=2000|num-b=14|num-a=16}}

Revision as of 01:23, 11 January 2009

Problem

Two non-zero real numbers, $a$ and $b$, satisfy $ab=a-b$. Find a possible value of $\frac{a}{b}+\frac{b}{a}-ab$.

$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -\frac{1}{2} \qquad\mathrm{(C)}\ \frac{1}{3} \qquad\mathrm{(D)}\ \frac{1}{2} \qquad\mathrm{(E)}\ 2$

Solution

$\begin{align*} \frac{a}{b}+\frac{b}{a}-ab &= \frac{a^2+b^2}{ab}-ab \\ &= \frac{-a^2b^2+a^2+b^2}{ab}\\ \end(align*}$ (Error compiling LaTeX. Unknown error_msg)

Substituting $ab=a-b$, we get

$\begin{align*} \frac{-(a+b)^2+a^2+b^2}{ab} &= \frac{-a^2+2ab-b^2+a^2+b^2}{ab} \\ &= \frac{2ab}{ab} \\ &= 2 \\ \end{align*}$ (Error compiling LaTeX. Unknown error_msg)

$\boxed{\text{E}}$

See Also

2000 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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
All AMC 10 Problems and Solutions