First, we solve the equation $\rm 3^{x-3} + 3^{4-x} = 4$ to find the possible values of x.

Given

$\rm 3^{x - 3} + 3^{4 - x} = 4$

By using the law of indices, $\rm a^{m-n} = \frac{a^{m}}{a^{n}}$,

$\rm or, \frac{3^{x}}{3^{3}} + \frac{3^{4}}{3^{x}} = 4$

$\rm or, \frac{3^{x}}{27} + \frac{81}{3^{x}} = 4$

Let $\rm 3^{x} = a$

$\rm or, \frac{a}{27} + \frac{81}{a} = 4$

Multiplying both sides of the equation by 27a, we get,

$\rm or, 27 a \cdot \frac{a}{27} + 27 a \cdot \frac{81}{a} = 27 a \cdot 4$

$\rm or, a^{2} + 2187 = 108 a$

$\rm or, a^{2} - 108 a + 2187 = 0$

The above equation is quadratic in a. We use the mid-term factorization method to solve it.

$\rm or, a^{2} - (27 + 81) a + 2187 = 0$

$\rm or, a^{2} - 27a - 81a + 2187 = 0$

$\rm or, a ( a - 27) - 81 ( a - 27) = 0$

$\rm or, (a - 81)(a - 27) = 0$

Either

$\rm (a - 81) = 0$

$\rm or, a = 81$

$\rm or, a = 3^{4}$

By supposition, $\rm a = 3^{x}$. We substitute the value for a and we get,

$\rm or, 3^{x} = 3^{4}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 4$

Or

$\rm (a - 27) = 0$

$\rm or, a = 27$

$\rm or, a = 3^{3}$

By supposition, $\rm a = 3^{x}$. We substitute the value for a and we get,

$\rm or, 3^{x} = 3^{3}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 3$

Hence, the possible values of x are x = {3,4}.

Finally, we substitute x = {3,4} in the second equation $\rm 4^{x-4} + 4^{3 - x} = 1 \frac{1}{4}$ to prove the condition.

Put x = 3

$\rm 4^{3 - 4} + 4^{3 - 3} = 1 \frac{1}{4}$

$\rm or, 4^{-1} + 4^{0} = \frac{4 \cdot 1 + 1}{4}$

By the law of indices $\rm a^{-m} = \frac{1}{a^{m}$ and $\rm a^{0} = 1$

$\rm or, \frac{1}{4} + 1 = \frac{4 + 1}{4}$

$\rm or, \frac{1 + 4}{4} = \frac{5}{4}$

$\rm \therefore \frac{5}{4} = \frac{5}{4}$ which is true.

Put x = 4

$\rm 4^{4 - 4} + 4^{3 - 4} = 1 \frac{1}{4}$

$\rm or, 4^{0} + 4^{-1} = \frac{ 4 \cdot 1 + 1}{4}$

By the law of indices $\rm a^{-m} = \frac{1}{a^{m}$ and $\rm a^{0} = 1$

$\rm or, 1 + \frac{1}{4} = \frac{4 + 1}{4}$

$\rm or,  \frac{ 4 + 1}{4} = \frac{5}{4}$

$\rm \therefore \frac{5}{4} = \frac{5}{4}$ which is true.

Hence, we conclude that the values of x in the equation $\rm 3^{x - 3} + 3^{4-x}$ also satisfy the equation $\rm 4^{x - 4} + 4^{3 - x} = 1 \frac{1}{4}$.