The perimeter of a rectangle is 28 ft. The length is 10 ft longer than the width. Find the dimensions. Write a system of linear equations and solve the resulting system. Let x be the length and y be the width.
To solve this problem, we need to set up a system of linear equations based on the given information.
Let's start by defining the variables:
x = length of the rectangle
y = width of the rectangle
Given that the length is 10 ft longer than the width, we can express the length in terms of the width:
x = y + 10
The perimeter of a rectangle is the sum of all four sides. In this case, the formula for the perimeter can be written as:
Perimeter = 2(length) + 2(width)
Since we are given that the perimeter is 28 ft, we can rewrite the equation as:
28 = 2(x) + 2(y)
Now, we have a system of two equations:
1) x = y + 10
2) 28 = 2x + 2y
To solve this system, we can use the substitution method or the elimination method. Let's use the substitution method:
Substitute the value of x from equation 1 into equation 2:
28 = 2(y + 10) + 2y
Simplify the equation:
28 = 2y + 20 + 2y
Combine like terms:
28 = 4y + 20
Subtract 20 from both sides:
8 = 4y
Divide both sides by 4 to solve for y:
y = 8/4
y = 2
Now, substitute the value of y back into equation 1 to solve for x:
x = y + 10
x = 2 + 10
x = 12
Therefore, the dimensions of the rectangle are:
Length (x) = 12 ft
Width (y) = 2 ft
oh drats non-linear is so much more fun!
where x is length and y is width.
2x + 2y = 28
x - 10 = y
solve with substitution