Mohammed had ducks and cows on his farm. These 14 animals had a total of 40 legs. How many each animals did Mohamed have?
assuming no amputees, there are 12 too many legs to be all ducks. So, 6 of the animals had 2 extra legs each.
8 ducks and 6 cows
or, more algebraically, if there are d ducks and c cows,
d+c = 14
2d+4c = 40
now solve for d and c.
Hi steve, thank you for this, if you please can you give me an explanation :( im short with answers, and quite busy at work.
Jahzarah Philips
Tfggfdgjgngfhfbfdbfbfcbfbgmhhjugjyy hhghfhfhhyfjgjugjugjujugjghgffhjy
To solve this problem, let's set up a system of equations using the given information:
Let's assume:
x = number of ducks Mohammed had
y = number of cows Mohammed had
Given that there were 14 animals in total, we can write the equation:
x + y = 14 (equation 1)
Furthermore, we know that the total number of legs is 40, and since ducks have 2 legs and cows have 4 legs, we can write a second equation:
2x + 4y = 40 (equation 2)
Now we have a system of two equations, and we can solve them simultaneously to find the values of x and y.
To solve the system, we can use a common method called substitution or elimination. Let's use substitution:
From equation 1, we can rearrange it to express x in terms of y:
x = 14 - y
Substituting the value of x in equation 2:
2(14 - y) + 4y = 40
28 - 2y + 4y = 40
2y = 40 - 28
2y = 12
y = 12 / 2
y = 6
Now that we know y (the number of cows), we can substitute it back into equation 1 to find x:
x + 6 = 14
x = 14 - 6
x = 8
Therefore, Mohammed had 8 ducks and 6 cows on his farm.