# 1970 AHSME Problems/Problem 26

## 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