Let $V$ be a vector space an choose $a,b,c\in V$ that are linearly dependent. We have $x,y,z\in\mathbb{F}$ (here $\mathbb{F}$ is the field associated with $V$). Thus \[ za=xb+yc. \] However; by algebra we have \[ a=\frac{x}{z}b+\frac{y}{z}c \Rightarrow a-\frac{x}{z}b-\frac{y}{z}c=0. \] Now we simply put $x_1=-x$, $y_1=-y$ and we multiply the entire equation by $z$, so \[ za-xb-yc=0 \] and finally we substitute in $x_1,y_1$ (which are in $\mathbb{F}$) to get \[ za+x_1b+y_1c=0 \]