The number of cattle at a farm is 660. This is a 10% decrease in the number of cows, a 50% increase in the number of bulls, and an overall increase of 10% in the total number of cattle from last year. How many cows and bulls were there each at the farm last year

let x = cows last year

let y = bulls last year

x + y = last years total

0.9x + 1.5y = 660 (this year's total)

1.1 (x+y) = 660 = 1.1x + 1.1y = 660

so:

0.9x + 1.5y = 1.1x + 1.1y

subtract 0.9x from each side:

1.5y = 0.2x + 1.1y

subtract 1.1y from each side

0.4y = 0.2x

to make x 1 multiply both sides by 5

2y = x

substitute for x in one of the original problems:

since they had 660 this year and that number is 10% higher than last year, then we had 600 total last year.

So:

600 = x + y

substitute into that one with the 2y = x

600 = 2y + y

600 = 3y

200 = y

then 600 - y = x

so 600 - 200 = x

400 = x

So you had 400 cows and 200 bulls last year