Precalculus (6th Edition) Blitzer

The coordinates of the line of the equation $2x+y=8$ are $\left( 1,6 \right),\left( 3,2 \right)$
Let us assume that x is the length of the rectangle and y is the breadth of the given rectangle. So, the first condition is the equation of the line written as: $2x+y=8$ (I) And the second condition is that the area of rectangle, which is $6$ square units, can be represented as the product of its length and its width as given below: $x\cdot y=6$ (II) Therefore, from equation (II), the value of y can be determined: $y=\frac{6}{x}$ Now, put $y=\frac{6}{x}$ in equation (I) and solve for the value of x: \begin{align} & 2x+y=8 \\ & 2x+\frac{6}{x}=8 \\ & 2{{x}^{2}}+6=8x \\ & 2{{x}^{2}}-8x+6=0 \end{align} Taking $2$ as a common factor from the above equation: \begin{align} & 2\left( {{x}^{2}}-4x+3 \right)=0 \\ & {{x}^{2}}-3x-x+3=0 \\ & x\left( x-3 \right)-1\left( x-3 \right)=0 \\ & \left( x-1 \right)\left( x-3 \right)=0 \end{align} This implies that $x=1$ and $x=3$. Now, substitute the values of x into $y=\frac{6}{x}$. When $x=1$, $y=6$. And when $x=3$, $y=2$ Thus, the coordinates of the line of equation $2x+y=8$ are $\left( 1,6 \right)$ and $\left( 3,2 \right)$.