Calculus (solve for $d_1$) I am learning Calculus on my own and came through this example and I solved the first two. However I couldn't find a solution for the last one.
Can you help ?
My solution is 
The first
 x =1
The second  x=2 & y=1



 A: $\epsilon = (d_2-d_1)/d_1$ Multiplying both sides by $d_1$, $\epsilon d_1=d_2-d_1$, adding $d_1 $ to both sides of the equation:  $\epsilon d_1+d_1=d_2$ factoring out $d_1$ we get: $d_1(\epsilon+1)=d_2$ And finally, we get: $d_1=(d_2)/(\epsilon+1)$. Note that I have no value given in the question for $d_2$ or $\epsilon$. However, this is the answer given the information that I have. 
A: They are asking you to solve $d_1$ in terms of $\epsilon$ and $d_2$, so the answer is not actually a number.
The first helpful thing to do here is to get rid of the fraction. This can be done by multiplying both sides by $d_1$:
$$d_1\epsilon=d_2-d_1$$
Now, since we are solving for $d_1$, we want to get all of the terms containing $d_1$ on the left side and all of the other terms on the right side. Therefore, add both sides by $d_1$:
$$d_1\epsilon+d_1=d_2$$
Now that all terms on the left side have $d_1$, we can factor out a $d_1$ from that expression:
$$d_1(\epsilon+1)=d_2$$
Now, finally, to isolate $d_1$, simply divide both sides by the expression $\epsilon+1$:
$$d_1=\frac{d_2}{\epsilon+1}$$
A: I'm not sure I would reccomend this method, but there is another way of solving ths equation. If you find it confusing feel free to ignore it, the ways in the other answers are better:
On the right hand side of the equal sign, you can split the fraction as follows:
$$\epsilon = \dfrac {d_2}{d_1} - \dfrac {d_1}{d_1}$$
Since $\dfrac {d_1}{d_1}=1$, we can write
$$\epsilon = \dfrac {d_2}{d_1} - 1$$
Adding $1$ to both sides, 
$$\epsilon +1 = \dfrac {d_2}{d_1} $$
From here you can multiply both sides by $d_1$, but don't distribute:
$$d_1 ( \epsilon + 1) = d_2$$
Now to islolate $d_1$, divide both sides by $\epsilon + 1$:
$$d_1 = \dfrac {d_2}{\epsilon + 1}$$
