Difference between revisions of "1970 AHSME Problems/Problem 26"
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\text{(F) } \infty</math> | \text{(F) } \infty</math> | ||
| − | == Solution == | + | == Solution 1== |
The graph <math>(x + y - 5)(2x - 3y + 5) = 0</math> is the combined graphs of <math>x + y - 5=0</math> and <math>2x - 3y + 5 = 0</math>. Likewise, the graph <math>(x -y + 1)(3x + 2y - 12) = 0</math> is the combined graphs of <math>x-y+1=0</math> and <math>3x+2y-12=0</math>. All these lines intersect at one point, <math>(2,3)</math>. | The graph <math>(x + y - 5)(2x - 3y + 5) = 0</math> is the combined graphs of <math>x + y - 5=0</math> and <math>2x - 3y + 5 = 0</math>. Likewise, the graph <math>(x -y + 1)(3x + 2y - 12) = 0</math> is the combined graphs of <math>x-y+1=0</math> and <math>3x+2y-12=0</math>. All these lines intersect at one point, <math>(2,3)</math>. | ||
Therefore, the answer is <math>\fbox{(B) 1}</math>. | Therefore, the answer is <math>\fbox{(B) 1}</math>. | ||
| + | == Solution 2== | ||
| + | We need to satisfy both <math>(x+y-5)(2x-3y+5)=0</math> and <math>(x-y+1)(3x+2y-12)=0</math>. In order to do this, let us look at the first equation. Either <math>x+y=5,</math> or <math>2x-3y=-5.</math> For the second equation, either <math>x-y=-1,</math> or <math>3x+2y=12.</math> Thus, we need to solve 4 systems of equations - <math>x+y=5</math> and <math>x-y=-1</math>, <math>x+y=5</math> and <math>3x+2y=12</math>, <math>2x-3y=-5</math> and <math>x-y=-1</math>, and finally <math>2x-3y=-5</math> and <math>3x+2y=12.</math> Solving all of these systems of equations is pretty trivial, and all of them come out to be <math>(2,3).</math> Thus, they only intersect at <math>1</math> point, and our answer is <math>\fbox(B) 1}</math>. | ||
| + | |||
| + | ~SirAppel | ||
== See also == | == See also == | ||
Revision as of 19:15, 27 August 2023
Contents
Problem
The number of distinct points in the 
-plane common to the graphs of 
and 
is
Solution 1
The graph 
is the combined graphs of 
and 
. Likewise, the graph 
is the combined graphs of 
and 
. All these lines intersect at one point, 
.
Therefore, the answer is 
.
Solution 2
We need to satisfy both and . In order to do this, let us look at the first equation. Either 
or 
For the second equation, either 
or 
Thus, we need to solve 4 systems of equations - 
and 
, and 
, 
and , and finally and Solving all of these systems of equations is pretty trivial, and all of them come out to be 
Thus, they only intersect at 
point, and our answer is $\fbox(B) 1}$ (Error compiling LaTeX. Unknown error_msg).
~SirAppel
See also
| 1970 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 25 |
Followed by Problem 27 | |
| 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. 
