Question 1011120
.
<pre>
Solve the four equations simultaneously

2a + c = 0          (1)
3a + 2b + d = 3     (2)
a + 3b + c = 2      (3)
b + d = -1          (4)
---------------------------------------

Express c = -2a from (1) and substitute it into (3). You will get

a + 3b - 2a = 3,   or   
-a + 3b = 2         (3')

Next, express d = -1 - b from(4) and substitute it into (2). You will get

3a + 2b + (-1 - b) = 3,   or
3a + b  = 4         (2').

Now you have this system of two equation in two unknowns (3') and (2').
To solve it, express b = 4 - 3a from (2') and substitute it into (3'). You will get

-a + 3*(4 - 3a) = 2  ----->  -a - 9a = 2 - 12  ----->  -10a = - 10  ----->  a = 1.

Then from (2')  b = 4 - 3a = 4 - 3 = 1.

Finally, from (1)  c = -2a = -2*1 = -2;
         from (4)  d = -1 - b = -1 - 1 = -2.

<U>Answer</U>. a = 1; b = 1; c = -2; d = -2.
</pre>