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 #2
avatar+118723 
+8
Jun 18, 2014
 #1
avatar+59 
0

To solve non-linear simultaneous equations, you must find the point at which they overlap. This can be done by finding the "x" and "y" values that apply to both equations.

Let's assume we have the following two simulatenous equations:

y = 3x

y = x + 12

We know that in each of these equations both "3x" (the same as 3 multiplied by x) and "x + 12" are equal to the variable of "y". Since the two equations are simultaneous, or occuring on the same graph, 3x and x + 12 must be equal. Thus, we can safely infer that:

3x = x + 12

Now we can simplify this equation using algebra. We can start by subtracting x from both sides to cancel out the "x" on the right side:

3x - x = x - x + 12

We can see that on the right side of the equation, "x - x" can be simplified to equal 0, so we can cancel that out to get:

3x - x = + 12

or...

3x - x = 12

Also, we can simplify "3x - x" to give an  answer of "2x":

2x = 12

To simplify what we have now, we can divide each side by 2:

2x / 2 12 / 2

We can see that "2x / 2" is equal to "x" and that "12 / 2" is equal to 6, so we have a final answer of:

x = 6

Now that we know "x", all we have to do is find "y". This is really easy: we can just plug it into one of our first equations: "y = 3x" or "y = x+12". We'll do "y = 3x" for the sake of simplicity:

y = 3(6)

We can simplify "3(6)" by multiplying 3 and 6 to obtain an answer of 18, giving us our answer for "y":

y = 18

Points on a graph are shown as (x,y), so through using algebraic methods, we were able to find that the two equations intersect at the point (6,18).

If I didn't help you, I found a really helpful Purplemath page that further explains non-linear simultaneous equations you can see here.

-pokemonfan58

Jun 18, 2014
 #3
avatar+163 
+8

The present value of a perpetuity-immediate (not perpetuity due) be C/i

The present value of a perpetuity-immediate of 10 is 10/i where i be the interest of land's userers. This be the interest for 1 annum, not 3. For 3, the effective  interest  be (i+1)^3 – 1.

Here the value of v is set to the inverse of

$$\ ( i+1): v=\frac{1}{(i+1)}.$$

The units of the Kingdom be Euros and marked as thus: €

$$v^{3}.\frac{10}{(i+1)^3-1}=32 \ Be\ the\ present\ value\ of\ the\ first\ perpetuity$$

$$\ Multiply \ by \ (i+1)^3 \ yieldeth \rightarrow \\
\hspace{150pt} \ 10v^3.(i+1)^3 = 32(i+1)^6 -32(i+1)^3$$

$$\\ In\ that\ v^3*\ (i+1)^3\ = \ 1; \rightarrow 10=32(i+1)^6-32(i+1)^3 \\
\ and \ becometh\ \rightarrow \ 0=32(i+1)^6-32(i+1)^3-10$$

With the magick of the quadratic there be a number that cometh forth. Aye! Two numbers, one be less than nothing. Sir CPhill was hither sent on a quest for nothing, and he findeth nothing, too, or not. This be less than nothing; I shall leave it for the gods, and taketh the weightier part.

$$\ (i+1)^3 = 32 \pm\frac {\sqrt{2304}}{64}=1.25$$

$$\ (i+1)^3=1.25 \
\\ and\ the\ user's\ rate\ be\
\\\sqrt[3]{1.25}-1= 0.0772 \ per\ \ annum\
\\ And\ for\ four\ moons\ be\ \sqrt[3]{1.0772} -1
\\The\ present\ value\ of\ the\ perpetuity-immediate \ be
\\\frac{1}{\sqrt[3]{1.0772} -1} \ =\ 39.84.\ In\ the\ units\ that\ be\ reinout-g's\ Euro$$

                                                                                                                   

Presented to and for her Royal Highness

by sevice of

Morgan Tud M-1

Jun 18, 2014
Jun 17, 2014

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