Answer
$d_{1}$ = 8i + 16j
Work Step by Step
$d_{1}$ + $d_{2}$ = 5$d_{3}$
$d_{1}$ - $d_{2}$ = 3$d_{3}$
$d_{3}$ = 2i + 4j
Solve by substitution
$d_{1}$ + $d_{2}$ = 5$d_{3}$
$d_{2}$ = 5$d_{3}$ - $d_{1}$
Now substitute the $d_{2}$ from above into the second given equation of $d_{1}$ - $d_{2}$ = 3$d_{3}$
Which gives you...
$d_{1}$ - 5$d_{3}$ + $d_{1}$ = 3$d_{3}$
Which reduces to
2$d_{1}$ = 8$d_{3}$
Which gives us
$d_{1}$ = 4$d_{3}$
Now we have to substitute in the third given equation of $d_{3}$ = 2i + 4j, so that $d_{1}$ will be in unit vector notation. Thus...
$d_{1}$ = 8i + 16j
