Answer
$x= \frac{1}{5}(4x'-3y')$ and $y= \frac{1}{5}(3x'+4y')$
Work Step by Step
1. Based on the given equation, we have $A=34, B=-24, C=41$, the rotation angle satisfies $cot(2\theta)=\frac{A-C}{B}=\frac{34-41}{-24}=\frac{7}{24}$ (quadrant I),
2. Let $x=7, y=24$, we have $r=25$, $cos(2\theta)=\frac{7}{25}$, with $\theta$ in quadrant I, thus $sin\theta=\sqrt {\frac{1-cos(2\theta)}{2}}=\sqrt {\frac{1-\frac{7}{25}}{2}}=\frac{3}{5}$ and
$cos\theta=\sqrt {\frac{1+cos(2\theta)}{2}}=\sqrt {\frac{1+\frac{7}{25}}{2}}=\frac{4}{5}$
3. The formulas for the rotation are $x=x'cos\theta-y'sin\theta=\frac{4}{5}x'-\frac{3}{5}y'=\frac{1}{5}(4x'-3y')$ and $y=x'sin\theta+y'cos\theta=\frac{3}{5}x'+\frac{4}{5}y'=\frac{1}{5}(3x'+4y')$
