Question

11 systems of equations: word problem practice objective: use the desmos graphing calculator to solve these real - world scenarios. for the first five, the equations are provided. for the last five, you must write the equations before graphing. the rectangular garden a gardener is fencing in a rectangular vegetable patch. the perimeter is 60 feet. the gardener wants the length (y) to be 4 feet longer than the width (x). what are the dimensions of the garden?

Explanation:

Step1: Define variables and relationships

Let the width be \( x \) and the length be \( y \). The length is 4 feet longer than the width, so \( y = x + 4 \). The perimeter of a rectangle is given by \( P = 2(x + y) \), and the perimeter is 60 feet, so \( 2(x + y)=60 \). Simplify the perimeter equation: divide both sides by 2, we get \( x + y = 30 \).

Step2: Substitute \( y \) into the perimeter equation

Substitute \( y = x + 4 \) into \( x + y = 30 \). So \( x+(x + 4)=30 \). Combine like terms: \( 2x+4 = 30 \). Subtract 4 from both sides: \( 2x=30 - 4=26 \). Divide both sides by 2: \( x=\frac{26}{2}=13 \).

Step3: Find the length \( y \)

Now that we know \( x = 13 \), substitute \( x \) into \( y = x + 4 \). So \( y=13 + 4 = 17 \).

Answer:

The width of the garden is 13 feet and the length is 17 feet.