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Difference between revisions of "1985 AHSME Problems/Problem 8"
(Improved wording, formatting, and LaTeX)
 
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==Problem==
 
==Problem==
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Let <math> a, a', b, </math> and <math> b' </math> be real numbers with <math> a </math> and <math> a' </math> nonzero. The solution to <math> ax+b=0 </math> is less than the solution to <math> a'x+b'=0 </math> if and only if  
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Let <math>a,a',b,b'</math> be real numbers with <math>a</math> and <math>a'</math> nonzero. The solution to <math>ax+b=0</math> is less than the solution to <math>a'x+b'=0</math> if and only if  
  
−
<math> \mathrm{(A)\ } a'b<ab' \qquad \mathrm{(B) \ }ab'<a'b \qquad \mathrm{(C) \  } ab<a'b' \qquad \mathrm{(D) \  } \frac{b}{a}<\frac{b'}{a'} \qquad </math>
+
<math> \mathrm{(A)\ } a'b < ab' \qquad \mathrm{(B) \ }ab' < a'b \qquad \mathrm{(C) \  } ab < a'b' \qquad \mathrm{(D) \  } \frac{b}{a} < \frac{b'}{a'} \qquad \mathrm{(E) \  }\frac{b'}{a'} < \frac{b}{a}  </math>
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<math> \mathrm{(E) \  }\frac{b'}{a'}<\frac{b}{a}  </math>
 
  
 
==Solution==
 
==Solution==
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The solution to <math> ax+b=0 </math> is <math> \frac{-b}{a} </math> and the solution to <math> a'x+b'=0 </math> is <math> \frac{-b'}{a'} </math>. The first solution is less than the second if <math> \frac{-b}{a}<\frac{-b'}{a'} </math>. We can get rid of the negative signs if we reverse the inequality sign, so we have <math> \frac{b'}{a'}<\frac{b}{a} </math>. All our steps are reversible, so our answer is <math> \boxed{\text{E}} </math>.
+
The solution to <math>ax+b=0</math> is <math>x = \frac{-b}{a}</math>, while that to <math>a'x+b'=0</math> is <math>x = \frac{-b'}{a'}</math>. The first solution is less than the second precisely if <cmath>\frac{-b}{a} < \frac{-b'}{a'},</cmath> and multiplying this inequality by <math>-1</math> reversees the inequality sign, yielding <math>\boxed{\text{(E)} \ \frac{b'}{a'} < \frac{b}{a}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=7|num-a=9}}
 
{{AHSME box|year=1985|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:13, 19 March 2024

Problem

Let $a,a',b,b'$ be real numbers with $a$ and $a'$ nonzero. The solution to $ax+b=0$ is less than the solution to $a'x+b'=0$ if and only if

$\mathrm{(A)\ } a'b < ab' \qquad \mathrm{(B) \ }ab' < a'b \qquad \mathrm{(C) \ } ab < a'b' \qquad \mathrm{(D) \ } \frac{b}{a} < \frac{b'}{a'} \qquad \mathrm{(E) \ }\frac{b'}{a'} < \frac{b}{a}$

Solution

The solution to $ax+b=0$ is $x = \frac{-b}{a}$, while that to $a'x+b'=0$ is $x = \frac{-b'}{a'}$. The first solution is less than the second precisely if \[\frac{-b}{a} < \frac{-b'}{a'},\] and multiplying this inequality by $-1$ reversees the inequality sign, yielding $\boxed{\text{(E)} \ \frac{b'}{a'} < \frac{b}{a}}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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