Question

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. which equation can be used to find y, the year in which both bodies of water have the same amount of mercury? o 0.05 - 0.1y = 0.12 - 0.06y o 0.05y + 0.1 = 0.12y + 0.06 o 0.05 + 0.1y = 0.12 + 0.06y o 0.05y - 0.1 = 0.12y - 0.06

Explanation:

Step1: Set up linear - equation for first body

The amount of mercury in the first body of water after \(y\) years is the initial amount plus the increase over \(y\) years. The initial amount is \(0.05\) ppb and the rate of increase is \(0.1\) ppb per year. So the amount of mercury \(M_1\) after \(y\) years is \(M_1=0.05 + 0.1y\).

Step2: Set up linear - equation for second body

The amount of mercury in the second body of water after \(y\) years is the initial amount plus the increase over \(y\) years. The initial amount is \(0.12\) ppb and the rate of increase is \(0.06\) ppb per year. So the amount of mercury \(M_2\) after \(y\) years is \(M_2 = 0.12+0.06y\).

Step3: Find the equation for equal amounts

We want to find when \(M_1 = M_2\). So we set \(0.05 + 0.1y=0.12 + 0.06y\).

Answer:

C. \(0.05 + 0.1y = 0.12+0.06y\)