Questions: The perimeter of a rectangle is 80 meters and the length is 10 meters longer than the width. Find the dimensions of the rectangle. Let x= the length and y= the width. The corresponding modeling system is 2x+2y=80, x-y=10. Solve the system graphically.

The perimeter of a rectangle is 80 meters and the length is 10 meters longer than the width. Find the dimensions of the rectangle. Let x= the length and y= the width. The corresponding modeling system is 2x+2y=80, x-y=10. Solve the system graphically.

Transcript text: The perimeter of a rectangle is 80 meters and the length is 10 meters longer than the width. Find the dimensions of the rectangle. Let $x=$ the length and $y=$ the width. The corresponding modeling system is $\left\{\begin{array}{r}2 x+2 y=80 \\ x-y=10\end{array}\right.$. Solve the system graphically.

Solution

Solution Steps

Step 1: Substitute and Solve for One Variable

Given the equation \(x - y = d\), we express \(x\) in terms of \(y\) and \(d\): \(x = y + d\).

Step 2: Substitute into the Perimeter Equation

Substituting \(x = y + d\) into the perimeter equation \(2x + 2y = P\) gives \(2(y + d) + 2y = P\), which simplifies to \(4y + 2d = P\).

Step 3: Solve for \(y\)

Rearranging the equation to solve for \(y\), we get \(y = \frac{{P - 2d}}{{4}} = 15\).

Step 4: Solve for \(x\)

Substituting the value of \(y\) back into the equation \(x = y + d\) to find \(x\), we get \(x = 25\).

Final Answer:

The dimensions of the rectangle are length \(x = 25\) and width \(y = 15\).

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