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If $x$ and $y$ are positive integers for which $3x + 2y + xy = 115$, then what is $x + y$?

 Jan 15, 2020
 #1
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x + y = 11.

 
 Jan 15, 2020
 #2
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If \(x\) and \( y\) are positive integers for which \(3x + 2y + xy = 115\), then what is \(x + y\)?

 

\(\begin{array}{|lrcll|} \hline & 3x + 2y + xy &=& 115 \quad | \quad + 3*2 \\ & 3x + 2y + xy + 3*2 &=& 115 + 3*2 \\ \text{First solution}& \mathbf{(x+2)(y+3)} &=& \mathbf{121} \\ \text{Second solution}& \mathbf{[-(x+2)][-(y+3)]} &=& \mathbf{121} \\ \hline \end{array}\)

 

\(\text{The divisors of $\mathbf{121}: 1 |\ 11 |\ 121$ }\\ \begin{array}{|rclcl|} \hline 1&*&121 &=& 121 \\ 11&*&11 &=& 121 \\ 121&*&1 &=& 121 \\ \hline \end{array} \)

 

\(\text{First solution}\quad \mathbf{(x+2)(y+3)} = \mathbf{121} \)

\(\begin{array}{|r|r|r|r|r|} \hline x+2 & y+3 & x & y & x+y \\ \hline 1 & 121 & -1 & 118 & 117 \\ 11 & 11 & \color{red}9 & \color{red}8 & \color{red}17 & \checkmark \text{ $x$ and $y$ are positive integers} \\ 121 & 1 & 119 & -2 & 117 \\ \hline \end{array} \)

 

\(\text{Second solution}\quad \mathbf{[-(x+2)][-(y+3)]} = \mathbf{121}\)

\(\begin{array}{|r|r|r|r|r|} \hline -(x+2) & -(y+3) & x & y & x+y \\ \hline 1 & 121 & -3 & -124 & -127 \\ 11 & 11 & -13 & -14 & -27 \\ 121 & 1 & -123 & -4 & -127 \\ \hline \end{array}\)

 

laugh

 
 Jan 15, 2020

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