Solving simultaneos equations

Question

I am new to simultaneous equations and need some help on this question:

Solve $$ \begin{align} x^{2 + 4y^2} &= 4 \\ y &= x - 1 \end{align} $$

Answer 1

OK, I think there are only numerical methods.

Substituting the 2nd expression into the first gives us

$$x^{2+4(x-1)^2} = 4$$

Taking $\ln$ on both sides gives us

$$(2+4(x-1)^2) \ln(x) - \ln(4) = 0$$

We now apply Newton-Raphson method. Let $f(x)$ denote the LHS. Thus,

$$f'(x) = 8(x-1)\ln(x) + \frac{1}{x} (2+4(x-1)^2)$$

For a starting value, let $x_0=1$. Thus,

$$\begin{align} x_1 &= x_0 - \frac{f(x_0)}{f'(x_0)}\\ &= 1 - \frac{f(1)}{f'(1)}\\ &= 1.693147181...\\ x_2 &= x_1 - \frac{f(x_1)}{f'(x_1)}\\ &= 1.693147181... - \frac{f(1.693147181...)}{f'(1.693147181...)}\\ &= 1.563497062...\\ x_3 &= x_2 - \frac{f(x_2)}{f'(x_2)}\\ &= 1.563497062... - \frac{f(1.563497062...)}{f'(1.563497062...)}\\ &= 1.545183059...\\ x_4 &= x_3 - \frac{f(x_3)}{f'(x_3)}\\ &= 1.545183059... - \frac{f(1.545183059...)}{f'(1.545183059...)}\\ &= 1.544847225...\\ x_5 &= x_4 - \frac{f(x_4)}{f'(x_4)}\\ &= 1.544847225... - \frac{f(1.544847225...)}{f'(1.544847225...)}\\ &= 1.544847114...\\ \end{align}$$

At this point, we have got a fairly good approximation for $x$, so we have $x=1.544847114...$, and $y=0.544847114...$

Answer 2

HINT: plugging the equation $$y=x-1$$ in the first equation we get $$x^{2+4(x-1)^2}=4$$ taking the logarithm on both sides we obtain: $$(1+2(x-1)^2)\ln(x)=\ln(2)$$ can you go on? one can also write $$x^{1+2y^2}=2$$ or $$x^{1+2y^2}=-2$$