Difference between revisions of "1960 AHSME Problems/Problem 5"

(Solution to Problem 5)
 
m (See Also)
 
(One intermediate revision by the same user not shown)
Line 24: Line 24:
 
==See Also==
 
==See Also==
  
{{AHSME box|year=1960|num-b=4|num-a=6}}
+
{{AHSME 40p box|year=1960|num-b=4|num-a=6}}
 +
 
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 18:55, 17 May 2018

Problem

The number of distinct points common to the graphs of $x^2+y^2=9$ and $y^2=9$ is:

$\textbf{(A) }\text{infinitely many}\qquad \textbf{(B) }\text{four}\qquad \textbf{(C) }\text{two}\qquad \textbf{(D) }\text{one}\qquad \textbf{(E) }\text{none}$

Solution

Solve the second equation by taking the square root of both sides. \[y^2=9\] \[y=\pm3\]

Solve the first equation by substituting $y^2$ into the first equation then solving for $x$. \[x^2+9=9\] \[x^2=0\] \[x=0\]

The two solutions are $(0,3)$ and $(0,-3)$, so the answer is $\boxed{\textbf{(C)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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