How can the equations be 15 MB each? Given the edit, the equation for h(7) is only 2582 characters. But if you want to solve for more unknowns than equations you need special constraints. For example $x^2+y^2+z^2=0$ is one equation in three unknowns, but has only one solution. This can be harder to detect in a system. You can introduce an error term into each equation, sum the squares and feed it to a multidimensional minimizer. Change each $a=b$ to $a=b-e_i$ and then say $error = \sum e_i^2$ and minimize $error$ over the variables. This is a numeric solution, not an algebraic one. If the equations are easy, this will work.
The underlying question here, I beleive, is what's called monotony, which basically means increasing or decreasing. Consider a function a real-valued function $f$, that takes real variables, with the property that: $$\text{For real numbers $x,y$ of a subset A} \\ x<y \Rightarrow f(x) < f(y)$$
Then we say that $f$ is increasing (in $\text{A}$) This happens, for example, with $f\mid f(x)=2x$. On the other hand, take $g \mid g(x)=x^2$, and the numbers $-3,2$. Obviously $-3<-2$, but $g(-3) \not< g(2)$. Actually if $x<0<y$, then $g(x)\not <g(y)$. This is why $$\sqrt{8x^{2}+22x+15} > 4x+3\not\Rightarrow \ \big( \sqrt{8x^{2}+22x+15} \big) ^{2} > \big(4x+3\big) ^{2},
$$ because $4x+3$ can be negative for, say $x=-2$, which is in the domain you have found that the square root is real (and therefore the expression makes sense). In fact, $4x+3$ is negative if and only if $x<\large\frac{-3}{4}$. Also notice that $x>\large\frac{-3}{4}\Rightarrow $ $x \ge \large\frac{-5}{4}$, and so, for all $x>\large\frac{-3}{4}$ $$\big( \sqrt{8x^{2}+22x+15} \big) ^{2} > \big(4x+3\big) ^{2}
$$
You can continue as before, taking into account that the result applies for the mentioned $x$'s. Say you obtain $x\in P\subset\mathbb R$ ($P$ some subset of real numbers).
For $x < \frac{-3}{4}$ (for $x < \frac{-3}{2}$ to be precise, because the expression is only defined here), the expression holds always. There the original inequality implies one of two situations: $$\left\{\begin{align} x &<\frac{-3}{2} \\ &\mathrm {or} \\ x&\in P\subset\mathbb R\end{align}\right.$$
This should be the final result, correct me if I'm wrong!
As for roots and exponents in general, or any function we "apply" to an inequality (I use quotations because really we are just taking into account the definition of increasing or decreasing, rather than "applying" the function), we just have to find the subsets where the function increases or decreases (decreasing has the same definition, replacing the second "$<$" with "$>$"). Then the procedure will yield, for each subset, a different subset of real numbers in which the inequality holds (could be empty sets). For example, for natural $n$, $h\mid h(x)=x^n$ is increasing in all of $\mathbb R$ if and only if $n$ is odd, and behaves as $x^2$ for even $n$. Try figuring out similar properties for positive rational exponents and positive $x$ (hint: write $x^{p/q}$ as $(x^p)^{\frac1q}$ and use the rules for natural exponents shown before, and think whether the $q$-th root of a number is increasing or decreasing (the function that assigns the $q$-th root). Hope I helped out!
Best Answer
One rather general strategy is to replace each new root $\sqrt[k]{expression}$ in the equation by a new variable, $r_j$, together with a new equation $r_j^k = expression$ (so now you will have $m+1$ polynomial equations in $m+1$ unknowns, where $m$ is the number of roots). Then eliminate variables from the system, ending with a single polynomial equation in one unknown, such that your original variable can be expressed in terms of the roots of this polynomial. This procedure can introduce spurious solutions if you only want the principal branch of the $k$'th root, so don't forget to check whether the solutions you get are valid.
For example, in your second equation, we get the system $$ \eqalign{r_1 + r_2 + r_3 - 2 \sqrt{2} &= 0\cr r_1^2-(3x-1) &= 0\cr r_2^2-(5x-3) &= 0\cr r_3^2-(x-1) &=0\cr}$$ Take the resultant of the first two polynomials with respect to $r_1$, then the resultant of this and the third with respect to $r_2$, and the resultant of this and the fourth with respect to $r_3$. We get $$ 121 x^4-4820 x^3+28646 x^2-45364 x+21417$$ which happens to factor as $$ \left( x-1 \right) \left( x-33 \right) \left( 121\,{x}^{2}-706\,x+ 649 \right) $$ However, only the solution $x=1$ turns out to satisfy the original equation.