Which equation can be used to find y, the year in which

both bodies of water have the same amount of mercury?

The levels of mercury in two different bodies of water are

rising. In one body of water the initial measure of mercury

is 0.05 parts per billion (ppb) and is rising at a rate of 0.1

ppb each year. In the second body of water the initial

measure is 0.12 ppb and the rate of increase is 0.06 ppb

each year.

A. 0.05 - 0.1y = 0.12 - 0.06y

B. 0.05y + 0.1 = 0.12y + 0.06

C. 0.05 + 0.1y = 0.12 + 0.06y

D 0.05y - 0.1 = 0.12y - 0.06

C. 0.05 + 0.1y = 0.12 + 0.06y

Step-by-step explanation:

According to the problem, both expression are equal because they have the same amount of mercury.

When the problem specifies an initial value, this means that value is not with the variable, because is an initial constant. When the problem specifies a rate, this is a coefficient of the variable, which expresses time (years.)

So, 0.05 parts per billion with a rate of 0.1 means: 0.05 + 0.1y, because the ratio is rising, that means is positive.

Similarly, 0.12 parts per billion with a rate of 0.06 means: 0.12 + 0.06y, because the ratio is also increasing, is positive.

Therefore, the right answer is C.