We first make a brief digression to look at equations from three other topics: $$\begin{align} x^3 + bx^2 + 3x - 2 &\equiv ax^3 + 4x^2 + 3x + c \tag{1} \\ 1 + b \sqrt{2} &= a + 4 \sqrt{2} \tag{2} \\ a + 4i &= 1 + bi \tag{3} \end{align} $$

For equation (1), when we work with identical polynomials, we can compare coefficients. This gives us $a=1, b=4$ and $c=-2$.
For equation (2), when working with surds, we can compare as well, provided $a$ and $b$ are integers/rational numbers.
Meanwhile, for equation (3), when working with complex numbers, we can compare the real and imaginary parts if $a$ and $b$ are real numbers.

Do discuss with me/your teacher if you are interested in why such comparisons work for each of the cases. As it turns out, a justification of all 3 cases can be explained at higher levels in (abstract) algebra in the study of rings and fields.

Now back to vectors. It turns out we can do a similar comparison procedure for two vectors $\mathbf{a}$ and $\mathbf{b}$, provided that $\mathbf{a}$ and $\mathbf{b}$ are non-parallel. This is illustrated in the example below, and is a very useful tool in solving questions when vectors are defined abstractly and not given in column vector form (or $\mathbf{ijk}$ notation).

Solution to example:

Since $O,X$ and $Y$ are collinear, $\overrightarrow{OX} = k\overrightarrow{OY}$.
$\lambda \mathbf{a} + \frac{1-\lambda}{2} \mathbf{b} = \frac{k}{2} \mathbf{a} + \frac{k}{2} \mathbf{b}$.
Comparing, since $\mathbf{a}$ and $\mathbf{b}$ are non-parallel, $\lambda = \frac{k}{2}$ and $\frac{1-\lambda}{2} = \frac{k}{2}$.
Solving the two equations simultaneously gives $\lambda = \frac{1}{3}$.