How do you evaluate #\frac { 3q + 1} { 3q ^ { 2} + 11q - 4} - \frac { 5q + 4} { 2q ^ { 2} + 7q - 4}#?

Your main concern is to make the denominators equal, since you're not going to be able to add or subtract fractions with contrasting denominators.

To do this, the first step is going to be to factor them.
#3q^2 + 11q - 4# can be rewritten as #(3q-1)(q+4)#, and you can check your work here by simplifying the factored form back to the original.
#(3q - 1)(q + 4)#
F #3q*q = 3q^2#
O #3q*4 = 12q#
I #-1 * q = -q#
L #-1 * 4 = -4#
#3q^2 + 12q - q - 4#
#3q^2 +11q - 4#
This checks out, so we're good here.

#2q^2 + 7q - 4# can be rewritten as #(2q-1)(q+4)#; again, we can check this:
#(2q-1)(q+4)#
F #2q*q=2q^2#
O #2q*4=8q#
I #-1*q = -q#
L #-1*4=-4#
#2q^2+8q-q-4#
#2q^2+7q-4#
Everything checks out again.

Now we can rewrite the expression as

#(3q+1)/((3q-1)(q+4)) - (5q+4)/((2q-1)(q+4))#

To make the denominators equal, we can multiply both sides of the first fraction by #(2q-1)#, the unique factor from the second fraction, and multiply both sides of the second fraction by #(3q-1)#, the unique factor from the first factor.

#((3q+1)(2q-1))/((3q-1)(q+4)(2q-1)) - ((5q+4)(3q-1))/((2q-1)(q+4)(3q-1))#

#((3q+1)(2q-1)-(5q+4)(3q-1))/((3q-1)(q+4)(2q-1))#

From here, all that's left to do is to simplify.

#((6q^2 - q - 1) - (15q^2 + 7q - 4))/((3q-1)(q+4)(2q-1))#

#(6q^2 - q - 1 - 15q^2 - 7q + 4)/((3q-1)(q+4)(2q-1))#

#(-9q^2 - 8q + 3)/((3q-1)(q+4)(2q-1))#

#(-9q^2 - 8q + 3)/((3q^2 + 11q - 4)(2q - 1))#

#(-9q^2 - 8q + 3)/(6q^3 + 22q^2 - 8q - 3q^2 - 11q + 4)#

Then we come to our final evaluation: #(-9q^2 - 8q + 3)/(6q^3 + 19q^2 - 19q + 4)#