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

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== Solution ==
 
== Solution ==
<math>\fbox{B}</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>.
  
 
== See also ==
 
== See also ==

Revision as of 22:41, 20 December 2017

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

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 $\fbox{(B) 1}$.

See also

1970 AHSC (ProblemsAnswer KeyResources)
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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