Respuesta :
Let the cost of a chicken be x and the cost of a duck, y, then
50x + 30y = 550 . . . (1)
44x + 36y = 532 . . . (2)
(1) x 6 ⇒ 300x + 180y = 3,300 . . . (3)
(2) x 5 ⇒ 220x + 180y = 2,660 . . . (4)
(3) - (4) ⇒ 80x = 640
⇒ x = 640 / 80 = 8
From (1): 50(8) + 30y = 550
⇒ 30y = 550 - 400 = 150
⇒ y = 150 / 30 = 5
Therefore, the price of a chicken is $8 and the price of a duck is $5.
50x + 30y = 550 . . . (1)
44x + 36y = 532 . . . (2)
(1) x 6 ⇒ 300x + 180y = 3,300 . . . (3)
(2) x 5 ⇒ 220x + 180y = 2,660 . . . (4)
(3) - (4) ⇒ 80x = 640
⇒ x = 640 / 80 = 8
From (1): 50(8) + 30y = 550
⇒ 30y = 550 - 400 = 150
⇒ y = 150 / 30 = 5
Therefore, the price of a chicken is $8 and the price of a duck is $5.
The price of a single chicken is $8 and the price of a single duck is $5 and this can be determine by forming the linear equation in two variables.
Given :
- Michael sold last month, 50 chickens and 30 ducks for $550.
- This month, he sold 44 chickens and 36 ducks for $532.
Let the price of single chicken be 'a' and the price of single duck be 'b'. Then the linear equation will be:
[tex]50a + 30b = 550[/tex] --- (1)
[tex]44a + 36b = 532[/tex] --- (2)
Now, solve the equation (1) for 'b'.
[tex]b = \dfrac{550-50a}{30}[/tex]
[tex]b = \dfrac{55}{3}-\dfrac{5a}{3}[/tex] ---- (3)
Now, put the value of 'b' in equation (2).
[tex]44a+36\left(\dfrac{55}{3}-\dfrac{5a}{3}\right) = 532[/tex]
[tex]44a+660-60a=532[/tex]
128 = 16a
a = 8
Now, put the value of a in equation (3).
[tex]b = \dfrac{55}{3}-\dfrac{5\times 8}{3}[/tex]
b = 5
The price of a single chicken is $8 and the price of a single duck is $5.
For more information, refer the link given below:
https://brainly.com/question/21835898