Difference between revisions of "1969 AHSME Problems/Problem 18"
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== Solution == | == Solution == | ||
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| + | By the [[Zero Product Property]], <math>x-y+2=0</math> or <math>3x+y-4=0</math> in the first equation and <math>x+y-2=0</math> or <math>2x-5y+7=0</math> in the second equation. Thus, from the first equation, <math>y = x+2</math> or <math>y =-3x+4</math>, and from the second equation, <math>y=-x+2</math> or <math>y = \frac{2}{5}x + \frac{7}{5}</math>. | ||
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| + | If a point is common to the two graphs, then the point must be in one of the lines in the first equation as well as one of the lines in the second equation. Since the slopes of the lines are different, none of the lines are parallel. Thus, there are <math>2 \cdot 2 = \boxed{\textbf{(B) } 4}</math> points of intersection in the two graphs. | ||
== See also == | == See also == | ||
Latest revision as of 19:08, 23 February 2024
Problem
The number of points common to the graphs of
is:
Solution
By the Zero Product Property, 
or 
in the first equation and 
or 
in the second equation. Thus, from the first equation, 
or 
, and from the second equation, 
or 
.
If a point is common to the two graphs, then the point must be in one of the lines in the first equation as well as one of the lines in the second equation. Since the slopes of the lines are different, none of the lines are parallel. Thus, there are 
points of intersection in the two graphs.
See also
| 1969 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
