Cfrgtkky

The variables: H, M, X,Y,Z

The five equations:
$$
7.6times{10^{-3}} =frac{X H}{M} tag{1}$$
$$6.2times{10^{-8}} =frac{Y H}{X} tag{2}$$
$$2.1times{10^{-13}} =frac{Z H}{Y} tag{3}$$
$$H^2 +0.3H = X H + 2Y H + 3ZH+10^{-14}tag{4} $$
$$0.3 = M + X + Y + Ztag{5} $$
Quistion : Derive equation containing just H variable.

This seems to be a chemical problem. When I saw it, I suspected possible traps because of the very small numbers and I rewrote the equations as
$$a =frac{X H}{M} tag{1}$$
$$b =frac{Y H}{X} tag{2}$$
$$c =frac{Z H}{Y} tag{3}$$
$$H^2 +frac 3 {10}H = X H + 2Y H + 3ZH+dtag{4} $$
$$frac 3 {10}= M + X + Y + Ztag{5} $$

In a first step, as @user10354138 commented, I used equations $(1,2,3,5)$ to express $X,Y,Z,M$ as functions of $H$. This leads to
$$X=frac{3 a H^2}{10 left(a b c+a b H+a H^2+H^3right)}qquad qquad Y=frac{3 a b H}{10 left(a b c+a b H+a H^2+H^3right)}$$
$$Z=frac{3 a b c}{10 left(a b c+a b H+a H^2+H^3right)}qquad qquad M=frac{3 H^3}{10 left(a b c+a b H+a H^2+H^3right)}$$
Now, replace these expressions in $(4)$, reduce to same denominator (which we shall ignore assuming it is non-zero) and put everything on the same side. We shall get
$$-10 a b c d-2 a b (3 c+5 d)H+a (b (10 c-3)-10 d)H^2+10 (a b-d)H^3+(10 a+3)
H^4+10 H^5=0tag 6$$ This is just a polynomial of degree $5$ which cannot be solved analytically and for which numerical methods will be required.

Replace $a,b,c,d$ by their values; you will get very ugly values for the coefficients.

If you want to solve $(6)$ for $H$, since we know the usual ranges of $H$, use inspection with $H=10^{-pH}$. Plotting the function as a function of $pH$, you will notice a very fast decreasing function with a solution close to $pH=4$. So, start Newton iterations using $pH_0=4$. You should get the following iterates
$$left(
begin{array}{cc}
n & pH_n \
0 & 4.00000 \
1 & 4.10601 \
2 & 4.21036 \
3 & 4.31193 \
4 & 4.40889 \
5 & 4.49815 \
6 & 4.57469 \
7 & 4.63121 \
8 & 4.66103 \
9 & 4.66846 \
10 & 4.66886
end{array}
right)$$ that is to say $H=10^{-4.66886}=0.0000214358$.

Edit

Plotting the function given by $(6)$ is not very good. You can have a much better perception using $H=10^{-pH}$ and plot, as a function of $pH$
$$f(pH)=frac{H^2 +frac 3 {10}H } {X H + 2Y H + 3ZH+d }-1tag 7$$ and look for its zero. Zoom once around the solution to get an estimate of $4.7$.

Solving $(7)$, using as before $pH_0=4$, Newton iterations would be
$$left(
begin{array}{cc}
n & pH_n \
0 & 4.00000 \
1 & 4.39558 \
2 & 4.63729 \
3 & 4.66879 \
4 & 4.66886
end{array}
right)$$ which is much faster than the previous one.