Real beams rarely carry one tidy point load. They carry floors, snow and their own weight, spread along the whole length, or they carry several loads at once. Both cases fall to the same two ideas you already own: the vertical forces balance, and the moments about any point balance.

Start with a uniformly distributed load, written $w$. It is a load per unit length, in kilonewtons per metre, and it runs the full span at constant intensity. Multiply by the span $L$ in metres and you have the total load, $wL$, in kilonewtons. The arrangement is symmetric, so neither support can claim more than the other: $R = \dfrac{wL}{2}$, read aloud R equals w L over two, where $R$ is the reaction at EACH support in kilonewtons. Read backwards it becomes a design rule. If a bearing is rated for $R$, the beam may carry $w = \dfrac{2R}{L}$.

Be clear about what did the work there. It was symmetry, not a formula. Break the symmetry and the half-and-half split goes with it, which is exactly what two point loads do. Name the parts: $P_1$ and $P_2$ are the two loads in kilonewtons, and the subscripts only say which load is which, 1 for the one nearer support A. $a_1$ and $a_2$ are their distances from support A in metres, both measured from the SAME end. $L$ is the span, and $R_A$ is the reaction at A, the one you are solving for.

Take moments about B, so that the unknown at B drops out, and each load turns about B on an arm of $L - a$: $R_A = \dfrac{P_1 (L - a_1) + P_2 (L - a_2)}{L}$, read aloud R-A equals P-one times L minus a-one, plus P-two times L minus a-two, all over L. Each load is weighted by its distance from the FAR support, the same rule as last lesson with one more term. Then the other reaction costs nothing: $R_B = P_1 + P_2 - R_A$, because the two supports between them hold up everything. Check that sum every time. It takes three seconds, and it catches the classic slip of taking moments about the wrong end.