Difference between revisions of "1970 AHSME Problems/Problem 26"
(→Solution 2) |
(→Solution 2) |
||
Line 14: | Line 14: | ||
Therefore, the answer is <math>\fbox{(B) 1}</math>. | Therefore, the answer is <math>\fbox{(B) 1}</math>. | ||
== Solution 2== | == 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 | + | 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 | ~SirAppel |
Latest revision as of 19:16, 27 August 2023
Contents
[hide]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 .
~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.