Substitute the values of x, y and z to the expression:
[tex]x=5,\ y=3,\ z=14\\\\\dfrac{5x-6y+20z}{4yz}=\dfrac{(5)(5)-(6)(3)+(20)(14)}{(4)(3)(14)}=\dfrac{25-18+280}{(12)(14)}\\\\=\dfrac{7+280}{168}=\dfrac{287}{168}=1\dfrac{119}{168}=1\dfrac{119:7}{168:7}=1\dfrac{17}{24}[/tex]
