1987 AHSME Problems/Problem 11: Difference between revisions
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\textbf{(C)}\ c<\frac{3}{2} \qquad | \textbf{(C)}\ c<\frac{3}{2} \qquad | ||
\textbf{(D)}\ 0<c<\frac{3}{2}\\ \qquad | \textbf{(D)}\ 0<c<\frac{3}{2}\\ \qquad | ||
\textbf{(E)}\ -1<c<\frac{3}{2} </math> | \textbf{(E)}\ -1<c<\frac{3}{2} </math> | ||
== Solution == | |||
We can easily solve the equations algebraically to deduce <math>x = \frac{5}{c+1}</math> and <math>y = \frac{3-2c}{c+1}</math>. Thus we firstly need <math>x > 0 \implies c + 1 > 0 \implies c > -1</math>. Now <math>y > 0</math> implies <math>\frac{3-2c}{c+1} > 0</math>, and since we now know that <math>c+1</math> must be <math>>0</math>, the inequality simply becomes <math>3-2c > 0 \implies 3 > 2c \implies c < \frac{3}{2}</math>. Thus we combine the inequalities <math>c > -1</math> and <math>c < \frac{3}{2}</math> to get <math>-1 < c < \frac{3}{2}</math>, which is answer <math>\boxed{\text{E}}</math>. | |||
== See also == | == See also == | ||
Latest revision as of 13:13, 1 March 2018
Problem
Let 
be a constant. The simultaneous equations
have a solution 
inside Quadrant I if and only if
Solution
We can easily solve the equations algebraically to deduce 
and 
. Thus we firstly need 
. Now 
implies 
, and since we now know that 
must be 
, the inequality simply becomes 
. Thus we combine the inequalities 
and 
to get 
, which is answer 
.
See also
| 1987 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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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