Solve for $r$ and a from systems-of-equations

Question

Can anybody calculate the value of $r$ and a from systems-of-equations given below?

$$\frac{a(1-r^7)}{1-r}=86$$

$$\frac{a(1-r^{10})}{1-r}=-682$$

Answer 1

Hint: We have $$a(1-r^7)=86(1-r)$$ and $$a(1-r^{10})=-682(1-r)$$. Dividing both equations, simplifying and factorizing we get

$$-2 (r-1) (r+2) \left(43 r^8-43 r^7+129 r^6+126 r^5+132 r^4+120 r^3+144 r^2+96 r+192\right)=0$$

Answer 2

Make the ratio of the two expressions to get $$\frac{1-r^{10}}{1-r^7}=-\frac{341}{43}$$ In absolute value, the rhs is large and then $r$ can be "large"; so may be, for the time being we could write $$\frac{1-r^{10}}{1-r^7}=-\frac{341}{43}\simeq \frac{r^{10}}{r^7}=r^3 $$ Use you calculator to get, as an approximation $$r_0=-\sqrt[3]{\frac{341}{43}}\approx -1.99417$$

If you dont want to jumpt to the obvious conclusion, now consider that you look for the zero of function $$f(r)=43(1-r^{10})+341(1-r^7)$$ and build the Taylor series around $r=r_0$; this would give $$f(r)=\left(-43 r_0^{10}-341 r_0^7+384\right)-\left(430 r_0^9+2387 r_0^6\right) (r-r_0)+O\left((r-r_0)^2\right)$$ Ignoring the higher order tems, solve for $r$ to get $$r=r_0-\frac{43 r_0^{10}+341 r_0^7-384}{r_0^6 \left(430 r_0^3+2387\right)}$$ You know the exact value of $r_0^3$, then the exact value of $r_0^6$; now approximate $r_0^7\simeq -2 r_0^6$ and $r_0^{10}\simeq-2 r_0^3\, r_0^6$ . This then gives $$r_1=r_0-\frac{236672}{39651821}=-2.00014$$

You do not need to solve a polynomial of degree $10$. Be practical !

Answer 3

Can anybody calculate the value of $r$ and a from systems-of-equations given below?

$$\frac{a(1-r^7)}{1-r}=86$$

$$\frac{a(1-r^{10})}{1-r}=-682$$

Cancel out $ 1 - r$ from the denominator by substitution.

$$a = 86\frac{1-r}{1-r^7}$$

Now substitute:

$$ \frac{86( 1-r^{10} )}{1-r^7} = -682 \Rightarrow {86 - 86r^{10}} = -682 + 682r^7$$

Let $t = r^7$

$$768 = 862t + 86r^3$$

Forms a Cubic Equation, which needs to be solved.