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

First write the equations in the slope-intercept form:


Line 1 passes through the points (-6,-6) and (-4,-3).


Line 2 passes through the points (0,3) and (2,6).


These are two identical lines so the system of equations has infinitely many solutions. The answer is D.