Hello anemone!
My solution:
If we multiply the first equation by 18, and the rest by 13, and then add them together, we obtain:
\(\displaystyle 135a+140b+145c+150d=9500\)
Dividing through by 5, we then get:
\(\displaystyle 27a+28b+29c+30d=1900\)
Hello MarkFL!Hello anemone!
My solution:
If we multiply the first equation by 18, and the rest by 13, and then add them together, we obtain:
\(\displaystyle 135a+140b+145c+150d=9500\)
Dividing through by 5, we then get:
\(\displaystyle 27a+28b+29c+30d=1900\)
Thanks, but...I left out the part where I explicitly solved a 4X4 linear system to determine what I needed to multiply the equations by so that I got what I needed. 
Hello MarkFL!
Thanks for participating and hey, your solution is smarter and neater than mine! Bravo, Mark!
My solution:
Adding those four equations yields
$10(a + b + c + d) = 630$ and this gives $a + b + c + d=63$.
But we're asked to evaluate $27a + 28b + 29c + 30d$, and $27a + 28b + 29c + 30d=30(a+b+c+d)-(3a+2b+c)$.
So now our effort should be focused on finding the value for $3a+2b+c$.
Since $4(63)=252=262-10$, we could write
$4(a+ b + c + d)=a + 2b + 3c + 4d-10$
$3a+ 2b + c=-10$
and therefore $27a + 28b + 29c + 30d=30(63)-(-10)=1900$.
A magician never shows how they do their tricks
