Difference between revisions of "1978 AHSME Problems/Problem 20"
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\textbf{(E) }-8 </math> | \textbf{(E) }-8 </math> | ||
| − | === | + | ==Solution== |
| + | |||
| + | Take the first two expressions (you can actually take any two expressions): <math>\frac{a+b-c}{c}=\frac{a-b+c}{b}</math>. | ||
| + | |||
| + | <math>\frac{a+b}{c}=\frac{a+c}{b}</math> | ||
| + | |||
| + | <math>ab+b^2=ac+c^2</math> | ||
| + | |||
| + | <math>a(b-c)+b^2-c^2=0</math> | ||
| + | |||
| + | <math>(a+b+c)(b-c)=0</math> | ||
| + | |||
| + | <math>\Rightarrow a+b+c=0</math> OR <math>b=c</math> | ||
| + | |||
| + | The first solution gives us <math>x=\frac{(-c)(-a)(-b)}{abc}=-1</math>. | ||
| + | |||
| + | The second solution gives us <math>a=b=c</math>, and <math>x=\frac{8a^3}{a^3}=8</math>, which is not negative, so this solution doesn't work. | ||
| + | |||
| + | Therefore, <math>x=1\Rightarrow\boxed{A}</math>. | ||
Revision as of 11:00, 3 July 2016
If 
are non-zero real numbers such that 
,
and 
, and 
, then 
equals
Solution
Take the first two expressions (you can actually take any two expressions): 
.
OR 
The first solution gives us 
.
The second solution gives us 
, and 
, which is not negative, so this solution doesn't work.
Therefore, 
.
