length of a rectangle is 5y more than twice its width and area of the rectangle is 42 yards ^2, find the dimensions
width --- x
length ---- 2x + 5 (assuming your 5y is a typo)
x(2x+5) = 42
2x^2 + 5x - 42 = 0
(x+6)(2x-7) = 0
x = -6 or x = 7/2 , but the width cannot be negative, so
the width is 7/2 or 3.5
the length is 12
check:
3.5(12) = 42
To find the dimensions of the rectangle, we'll first set up an equation based on the information given.
Let's assume the width of the rectangle is 'w' yards. Then, according to the given information, the length of the rectangle would be '2w + 5y' yards.
We also know that the area of a rectangle is equal to its length multiplied by its width. In this case, the area is given as 42 yards^2. So we can write the equation as:
Area = Length x Width
42 = (2w + 5y) x w
Now, let's solve the equation to find the values of 'w' and 'y'.
42 = 2w^2 + 5yw
Rearranging the equation:
2w^2 + 5yw - 42 = 0
We now have a quadratic equation in terms of 'w'. We can solve this equation by factoring, completing the square, or using the quadratic formula.
Let's use the quadratic formula:
w = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, 'a' is the coefficient of w^2, 'b' is the coefficient of w, and 'c' is the constant term.
For our equation: 2w^2 + 5yw - 42 = 0,
a = 2, b = 5y, and c = -42.
Substituting these values into the quadratic formula:
w = (-5y ± √((5y)^2 - 4 * 2 * -42)) / (2 * 2)
Simplifying further:
w = (-5y ± √(25y^2 + 336)) / 4
Now, we have the value of 'w' in terms of 'y'. To find the exact dimensions, we'll need specific values of 'y'. If you have a specific value for 'y', you can substitute it into the above equation to find the corresponding value of 'w'.