Solve for a, b, c, d?

a + b = (c + d) / 2

a + c = 2b + 2d

a + d = 1.5b + 1.5c

1 Answer
Feb 2, 2018

#(a, b, c, d) = (9lambda, lambda, 11lambda, 9lambda)#

Explanation:

Multiplying the first and third equations by #2# and rearranging slightly, we have:

#{ (2a+2b-c-d = 0), (a-2b+c-2d=0), (2a-3b-3c+2d=0) :}#

Adding the first two equations, we get:

#3a-3d=0#

Hence:

#a = d#

Substituting #a# for #d# in the first and third equations, we get:

#{ (a+2b-c=0), (4a-3b-3c=0) :}#

Mutiplying the first equation by #3# we get:

#{ (3a+6b-3c=0), (4a-3b-3c=0) :}#

Subtracting the first of these from the second, we get:

#a-9b=0#

Hence:

#a = 9b#

From a previous equation we have:

#c = a+2b = 9b+2b = 11b#

Writing #b = lambda#, we find there are infinitely many solutions all taking the form:

#(a, b, c, d) = (9lambda, lambda, 11lambda, 9lambda)#