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Difference between revisions of "1964 AHSME Problems/Problem 18"

(Created page with "== Problem 18== Let <math>n</math> be the number of pairs of values of <math>b</math> and <math>c</math> such that <math>3x+by+c=0</math> and <math>cx-2y+12=0</math> have the...")
 
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The intercept of the first line is <math>\frac{-c}{b}</math>, while the intercept of the second line is <math>6</math>.  Thus, <math>6 = \frac{-c}{b}</math>, or <math>-6b = c</math>.
 
The intercept of the first line is <math>\frac{-c}{b}</math>, while the intercept of the second line is <math>6</math>.  Thus, <math>6 = \frac{-c}{b}</math>, or <math>-6b = c</math>.
  
−
Plugging <math>-6b = c</math> into <math>bc = -6</math> gives <math>b(-6b) = -6</math>, or <math>b^2 = 1</math>.  This means <math>b = \pm 1</math>  This in turn gives <math>c = \mp 6</math>.  Thus, <math>(b, c) = (\pm 1, \mp 6)</math>, for two solutions, which is answer <math>\boxed{\textbf{(D)}}</math>
+
Plugging <math>-6b = c</math> into <math>bc = -6</math> gives <math>b(-6b) = -6</math>, or <math>b^2 = 1</math>.  This means <math>b = \pm 1</math>  This in turn gives <math>c = \mp 6</math>.  Thus, <math>(b, c) = (\pm 1, \mp 6)</math>, for two solutions, which is answer <math>\boxed{\textbf{(C)}}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 11:25, 8 May 2020

Problem 18

Let $n$ be the number of pairs of values of $b$ and $c$ such that $3x+by+c=0$ and $cx-2y+12=0$ have the same graph. Then $n$ is:

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ \text{finite but more than 2}\qquad \textbf{(E)}\ \infty$

Solution

For two lines to be the same, their slopes must be equal and their intercepts must be equal. This is a necessary and sufficient condition.

The slope of the first line is $\frac{-3}{b}$, while the slope of the second line is $\frac{c}{2}$. Thus, $\frac{-3}{b} = \frac{c}{2}$, or $bc = -6$.

The intercept of the first line is $\frac{-c}{b}$, while the intercept of the second line is $6$. Thus, $6 = \frac{-c}{b}$, or $-6b = c$.

Plugging $-6b = c$ into $bc = -6$ gives $b(-6b) = -6$, or $b^2 = 1$. This means $b = \pm 1$ This in turn gives $c = \mp 6$. Thus, $(b, c) = (\pm 1, \mp 6)$, for two solutions, which is answer $\boxed{\textbf{(C)}}$

See Also

1964 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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