Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 8"

Latest revision as of 16:35, 6 May 2007

If we subtract from $2$ the inverse number of $x-1$ , we get the inverse of $x-1$ . Then the number $x+1$ equals to

$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } -1\qquad \mathrm{(D) \ } 3\qquad \mathrm{(E) \ } \frac{1}{2}$

$2-\frac1{x-1}=\frac1{x-1}$

$2=\frac2{x-1}$

Multiplying out and solving, we get that $x = 2$ , so $x+1=3\Longrightarrow\mathrm{ D}$ .

@@ Line 1: / Line 1: @@
 ==Problem==
-If we substract from <math>2</math> the inverse number of <math>x-1</math>, we get the inverse of <math>x-1</math>. Then the number <math>x+1</math> equals to
+If we subtract from <math>2</math> the [[reciprocal|inverse]] number of <math>x-1</math>, we get the inverse of <math>x-1</math>. Then the number <math>x+1</math> equals to
 <math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } -1\qquad \mathrm{(D) \ } 3\qquad \mathrm{(E) \ } \frac{1}{2}</math>
 ==Solution==
-<math>2-\frac1{x-1}=\frac1{x-1}</math>
+<div style="text-align:center;"><math>2-\frac1{x-1}=\frac1{x-1}</math>
+<br>
+<math>2=\frac2{x-1}</math></div>
-<math>2=\frac2{x-1}</math>
+Multiplying out and solving, we get that <math>x = 2</math>, so <math>x+1=3\Longrightarrow\mathrm{ D}</math>.
-<math>2x-2=2</math>
-<math>2x=4</math>
-<math>x=2</math>
-<math>x+1=3\Rightarrow\mathrm{ D}</math>
 ==See also==
-*[[2007 Cyprus MO/Lyceum/Problems]]
+{{CYMO box|year=2007|l=Lyceum|num-b=7|num-a=9}}
-*[[2007 Cyprus MO/Lyceum/Problem 7|Previous Problem]]
-*[[2007 Cyprus MO/Lyceum/Problem 9|Next Problem]]
+[[Category:Introductory Algebra Problems]]