Question

TThe area of a rectangle is 77 ft2, and the length of the rectangle is 10FT less than three times the width. Find the dimensions of the rectangle.

Answer 1

Call a the length and b the width'
a* b = 77 (1)
a + 10 = 3b -> a = (3b - 10). Replace this value of a into equation (1)
(3b - 10)* b = 77 -> 3b^2 - 10 b - 77 = 0. Solve this quadratic formula by the formula. D = b^2 - 4ac = 100 + 924 = 1024 = (32)^2
2 real roots: x1 = 10/6 + 32/6 = 42/6 = 7 = b
x2 = 10/6 - 32/6: rejected because < 0.
Finally: b = 7 -> a = 3b - 10 = 21 - 10 = ....

Answer 2

Thank you , but what is the answer. Im looking for the length and width

Answer 3

area of a rectangle = Length, l * w idth, w
L*W = 77
length =10 less than 3 times width
L = 3W - 10
you have 2 equations
substitute for L in the first to get one equation with one unknown, W
Solve for W
now that you have W put this value in the expression for L to get the value of L

always check your answers
L*W = 77

Answer 4

a is the length, and b is the width
b = 7 and a = .....

Answer 5

(3w-10)(w) = 77
3w^2-10W - 77 = 0
solve for the width using the discriminant above or by factoring
(3w + 11)(w-7)=0
w cannot be negative therefore correct width is 7
L = 3*7 -10 = 11
are you ok now?