Can you find the identity of ax + by?

We need to generate the $ax$ and $by$ terms for the solution. There are many ways to do that, but the requested one is to square the first two equations:

$(3a+4x)^2 = 6^2 \Rightarrow 9a^2+24ax+16x^2 = 36$

$(3b+4y)^2 = 1^2 \Rightarrow 9b^2+24by+16y^2 = 1$

Adding up the two results:

$9(a^2+b^2)+24(ax+by)+16(x^2+y^2)=37$

and substituing the other two equations:

$9+24(ax+by)+16=37 \Rightarrow 25+24(ax+by)=37 \Rightarrow 24(ax+by)=12$

Then, $ax+by=\frac {1}{2}$

Manrique's solution is certainly more elegant than mine was, but here goes:

$3a+4y = 6$

Add $3b+4y = 1$ into the first equation, one would get:

$3(a+b)+4(x+y) = 7$

Squaring and expanding, one would yield (ive skipped a few steps, but I think the general idea is there) :

$9(a^2+b^2)+16(x^2 + y^2) +24(ax + by) + 6a(3b+4y) + 8x(3b+4y) = 49$

Since $a^2 + b^2 = 1$ , $x^2+y^2 = 1$ , $3b+4y=1$ , $3a+4x = 6$ :

$24(ax+by) = 49 - 16 - 9 - 2(6)$

$ax + by = \frac{12}{24} = 0.5$