\(=4\cdot\dfrac{4}{3}+\dfrac{3}{35}=\dfrac{16}{3}+\dfrac{3}{35}=\dfrac{560}{105}+\dfrac{9}{105}=\dfrac{569}{105}\)
\(4\times\dfrac{4}{3}+\dfrac{1}{5}\times\dfrac{3}{7}=\dfrac{16}{3}+\dfrac{3}{35}=\dfrac{569}{105}\)
\(=\dfrac{16}{3}+\dfrac{3}{35}=\dfrac{560}{105}+\dfrac{9}{105}=\dfrac{569}{105}\)