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Difference between revisions of "1970 AHSME Problems/Problem 26"

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\text{(F) } \infty</math>
 
\text{(F) } \infty</math>
  
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== Solution ==
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== Solution 1==
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<math>\fbox{E}</math>
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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>.
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Therefore, the answer is <math>\fbox{(B) 1}</math>.
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== Solution 2==
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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>.
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~SirAppel
  
 
== See also ==
 
== See also ==

Latest revision as of 19:16, 27 August 2023

Problem

The number of distinct points in the $xy$-plane common to the graphs of $(x+y-5)(2x-3y+5)=0$ and $(x-y+1)(3x+2y-12)=0$ is

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) } 4\quad \text{(F) } \infty$

Solution 1

The graph $(x + y - 5)(2x - 3y + 5) = 0$ is the combined graphs of $x + y - 5=0$ and $2x - 3y + 5 = 0$. Likewise, the graph $(x -y + 1)(3x + 2y - 12) = 0$ is the combined graphs of $x-y+1=0$ and $3x+2y-12=0$. All these lines intersect at one point, $(2,3)$. Therefore, the answer is $\fbox{(B) 1}$.

Solution 2

We need to satisfy both $(x+y-5)(2x-3y+5)=0$ and $(x-y+1)(3x+2y-12)=0$. In order to do this, let us look at the first equation. Either $x+y=5,$ or $2x-3y=-5.$ For the second equation, either $x-y=-1,$ or $3x+2y=12.$ Thus, we need to solve 4 systems of equations: $x+y=5$ and $x-y=-1$, $x+y=5$ and $3x+2y=12$, $2x-3y=-5$ and $x-y=-1$, and finally $2x-3y=-5$ and $3x+2y=12.$ Solving all of these systems of equations is pretty trivial, and all of them come out to be $(2,3).$ Thus, they only intersect at $1$ point, and our answer is $\fbox{(B) 1}$.

~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

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