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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

The exercise

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closed as unclear what you're asking by Jack, Mark Viola, Shailesh, Daniel W. Farlow, user223391 Dec 19 '16 at 4:20

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    $\begingroup$ You might be used to "solving" a variables by finding an exact numerical value for it, but in algebra, sometimes you will have to find a value in terms of other variables. You might not be able to say what exact number $d_1$ is, but you can express it in terms of $\epsilon$ and $d_2$. $\endgroup$ – Kevin Long Dec 18 '16 at 19:59
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$\epsilon = (d_2-d_1)/d_1$ 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.

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  • $\begingroup$ Many thanks for your answer, do you think my solution for the first and the second is correct ? $\endgroup$ – Kkk Dec 18 '16 at 20:05
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    $\begingroup$ @Karla Yes, your solution to the second one is correct. You can check it by plugging in your values of $x$ and $y$ into the equations: $$2 \cdot (2) + 1 = 5$$ $$ 3 \cdot 2 + 4 \cdot 1 = 10$$ Since the left and right side of the equations agree for $both$ equations, your solution is correct. $\endgroup$ – Ovi Dec 18 '16 at 20:09
  • $\begingroup$ Ok, let's check. If we plug in $x=1$ into the first equation we get: $1+(-3)^2=10$ so we get, $10=10$, so the first answer checks out. For the second, note what @Ovi said $\endgroup$ – wesssg Dec 18 '16 at 20:10
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    $\begingroup$ @Karla, in your example for part b) you have the correct solution. However, In some system of equations there may be 1 solution, infinite solutions, or none. $\endgroup$ – wesssg Dec 18 '16 at 20:11
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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 $\epsilon+1$: $$d_1=\frac{d_2}{\epsilon+1}$$

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  • $\begingroup$ Many thanks for your answer. $\endgroup$ – Kkk Dec 18 '16 at 20:04
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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 $\epsilon + 1$:

$$d_1 = \dfrac {d_2}{\epsilon + 1}$$

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  • $\begingroup$ Many thanks for your answer. $\endgroup$ – Kkk Dec 18 '16 at 20:07
  • $\begingroup$ @Karla You're welcome $\endgroup$ – Ovi Dec 18 '16 at 20:11

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