The slaughterhouse brought a number of sheep at sh. 1200 each and a number of oxen at sh.15, 000 each.

Question

They paid a total of sh.135, 000. If they had brought twice as many sheep and three oxen less. They
would have saved sh. 15000.
a) Write two simultaneous equations presently the information.
b) Find the number of each type of animal they brought

Answer 1

Let's let the number of sheep be represented by "s" and the number of oxen be represented by "o".

a) The first equation can be formed from the given information that they paid sh.1200 for each sheep and sh.15000 for each oxen, and the total amount paid was sh.135000.

Therefore, the first equation is:
1200s + 15000o = 135000.

The second equation can be formed from the given information that if they had brought twice as many sheep (2s) and three fewer oxen (o-3), they would have saved sh.15000.

Therefore, the second equation is:
1200(2s) + 15000(o-3) = 135000 + 15000.

Simplifying the second equation:
2400s + 15000o - 45000 = 150000.

b) To find the number of each type of animal they brought, we can solve the two simultaneous equations.

1200s + 15000o = 135000 (Equation 1)
2400s + 15000o - 45000 = 150000 (Equation 2)

To solve the equations, we can use the elimination method. Subtracting Equation 1 from Equation 2, we get:

1200s - 45000 = 150000 - 135000,
which simplifies to:
1200s - 45000 = 15000.

Adding 45000 to both sides of the equation:
1200s = 60000.

Dividing both sides by 1200:
s = 50.

Now we can substitute the value of s into Equation 1 to find the value of o:

1200(50) + 15000o = 135000,
60000 + 15000o = 135000,
15000o = 135000 - 60000,
15000o = 75000.

Dividing both sides by 15000:
o = 5.

Therefore, they brought 50 sheep and 5 oxen.