Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Review Exercises - Page 877: 37


The coordinates of the line of the equation $2x+y=8$ are $\left( 1,6 \right),\left( 3,2 \right)$

Work Step by Step

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 $\left( 3,2 \right)$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.