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{3a+4x=63b+4y=1a2+b2=1x2+y2=1 \large { \begin{cases} {3a +4x = 6} \\ { 3b + 4y = 1 } \\ { a^2 + b^2= 1 } \\ {x^2 + y^2 = 1 } \end{cases} } ⎩⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎧3a+4x=63b+4y=1a2+b2=1x2+y2=1
Given that a,b,xa,b,xa,b,x and yyy satisfy the system of equations above, find the value of ax+byax + by ax+by.
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