Chapter 1 - Section 1.2 - The Geometry of Linear Equations - Problem Set - Page 9: 2

$$w =b_{3} $$ $$ v= b_{2}- b_{3}$$ $$u= b_{1}+ b_{2}$$

Given $$\begin{aligned} u-v-w &=b_{1} \\ v+w &=b_{2} \\ w &=b_{3} \end{aligned}$$ So, we have $$w =b_{3} $$ Since we have $$ v+w =b_{2}\\ \Rightarrow v=b_{2}-w= b_{2}- b_{3}$$ Also $$ u-v-w =b_{1}\\ \Rightarrow u=b_{1}+v+w= b_{1}+ b_{2}- b_{3}+ b_{3}\\ \Rightarrow u= b_{1}+ b_{2}$$