In 1979, zoologists estimated the number of fish in a local pond to be about 540. In 1989, 10 years later, the zoologists estimated that the number of
fish had increased to 990. Assuming that the fish population increased
linearly, how many fish lived in the pond in 1986? *
ann

693 fish

Step-by-step explanation:

In 10 years, we can tell that the amount of fish increased by 1.833...(990 divided by 540) times. If we divide 1.833... by 10, we can find how much the pond increases by yearly. we get 0.1833... If we multiply that by 7 (because the gap between 1986 and 1979 is 7 years), we get 1.2833... So, if we multiply 540 by 1.28333..., we get 693 fish.

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1