Difference between revisions of "1960 AHSME Problems/Problem 5"
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==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}} |
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| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 17:55, 17 May 2018
Problem
The number of distinct points common to the graphs of 
and 
is:
Solution
Solve the second equation by taking the square root of both sides.
![\[y^2=9\]](http://latex.artofproblemsolving.com/7/1/5/715fa800dddcafe15e3bd2137c512bb4f2c633e3.png)
![\[y=\pm3\]](http://latex.artofproblemsolving.com/9/e/9/9e9acc628d38bf809f81fd5e72278e8e4733e8bb.png)
Solve the first equation by substituting 
into the first equation then solving for 
.
![\[x^2+9=9\]](http://latex.artofproblemsolving.com/6/7/6/676c5b1c1338f5018af042f659e6645f68906e85.png)
![\[x^2=0\]](http://latex.artofproblemsolving.com/2/5/d/25d30d448342ef96ce0ef6fcd5531c1362d6b6d4.png)
![\[x=0\]](http://latex.artofproblemsolving.com/0/e/5/0e5065ea42e71241d1571fd868bae3267c53d818.png)
The two solutions are 
and 
, so the answer is 
.
See Also
| 1960 AHSC (Problems • Answer Key • Resources) | ||
| 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 | ||
