Triangle ABC is translated to triangle A'B'C' by the following motion rule. (x,y)→(x+2,y−5)
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In short, to convert two fractions to have the same denominator, we simply multiply one by the denominator of the other, so in this case, we'll multiply 1/3 by 5, top and bottom, and 1/5 by 3, top and bottom, thus
![\bf a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} \qquad \qquad \sqrt[ m]{a^ n}\implies a^{\frac{ n}{ m}}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20a%5E%7B%5Cfrac%7B%20n%7D%7B%20m%7D%7D%20%5Cimplies%20%20%5Csqrt%5B%20m%5D%7Ba%5E%20n%7D%20%0A%5Cqquad%20%5Cqquad%0A%5Csqrt%5B%20m%5D%7Ba%5E%20n%7D%5Cimplies%20a%5E%7B%5Cfrac%7B%20n%7D%7B%20m%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf \cfrac{15x^{\frac{1}{3}}}{y^{\frac{1}{5}}}\qquad \begin{cases} \frac{1}{3}=\frac{1\cdot 5}{3\cdot 5}\\ \qquad \frac{5}{15}\\\\ \frac{1}{5}=\frac{1\cdot 3}{5\cdot 3}\\ \qquad \frac{3}{15} \end{cases}\implies \cfrac{15x^{\frac{5}{15}}}{y^{\frac{3}{15}}}\implies 15\cdot \cfrac{x^{\frac{5}{15}}}{y^{\frac{3}{15}}}\implies 15\cdot \cfrac{\sqrt[15]{x^5}}{\sqrt[15]{y^3}} \\\\\\ 15\sqrt[15]{\frac{x^5}{y^3}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B15x%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D%7By%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0A%5Cfrac%7B1%7D%7B3%7D%3D%5Cfrac%7B1%5Ccdot%205%7D%7B3%5Ccdot%205%7D%5C%5C%0A%5Cqquad%20%5Cfrac%7B5%7D%7B15%7D%5C%5C%5C%5C%0A%5Cfrac%7B1%7D%7B5%7D%3D%5Cfrac%7B1%5Ccdot%203%7D%7B5%5Ccdot%203%7D%5C%5C%0A%5Cqquad%20%5Cfrac%7B3%7D%7B15%7D%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B15x%5E%7B%5Cfrac%7B5%7D%7B15%7D%7D%7D%7By%5E%7B%5Cfrac%7B3%7D%7B15%7D%7D%7D%5Cimplies%2015%5Ccdot%20%5Ccfrac%7Bx%5E%7B%5Cfrac%7B5%7D%7B15%7D%7D%7D%7By%5E%7B%5Cfrac%7B3%7D%7B15%7D%7D%7D%5Cimplies%2015%5Ccdot%20%5Ccfrac%7B%5Csqrt%5B15%5D%7Bx%5E5%7D%7D%7B%5Csqrt%5B15%5D%7By%5E3%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A15%5Csqrt%5B15%5D%7B%5Cfrac%7Bx%5E5%7D%7By%5E3%7D%7D)